Physics

Solved problems in vectors – determine resultant of two vectors using the Pythagorean theorem

1. Determine the resultant of the two displacement vectors as shown in the figure below.

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2. Find the resultant of two forces, 12 N and 5 N.

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3. A student walks 4 meters to the west, then 6 meters to the north and 4 meters to the west. Find the student displacement.

Solution

Displacement is 10 meter, to the northwest.

[irp]

[wpdm_package id=’542′]

[wpdm_package id=’554′]

1. Determine the resultant of in a line vector
2. Determine vector components
3. Determine the resultant of two vectors using the Pythagorean theorem
4. Determine the resultant of two vectors using cosines equation
5. Determine the resultant of two vectors using components of vectors

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Solved problems in vectors – determine vector components

1. A force of 20 Newton makes an angle of 30^o with the x-axis. Find both the x and y component of the force.

Solution

F_x = F cos 30^o = (20)(cos 30^o) = (20)(0.5√3) = 10√3 Newton

F_y = F sin 30^o = (20)(sin 30^o) = (20)(0.5) = 10 Newton

[irp]

2. F₁ = 20 Newton makes an angle of 30^o with the y axis and F₂ = 30 Newton makes an angle of 60^o with the -x axis. Find both the x and y components of F₁ and F₂.

Solution

F_1x = F₁ cos 60^o = (20)(cos 60^o) = (20)(0.5) = -10 Newton (negative because it has same direction with -x axis)

F_2x = F₂ cos 60^o = (30)(cos 60^o) = (30)(0.5) = -15 Newton (negative because it has same direction with -x axis)

F_1y = F₁ sin 60^o = (20)(sin 60^o) = (20)(0.5√3) = 10√3 Newton (positive because it has same direction with y axis)

F_2y = F₂ sin 60^o = (30)(sin 60^o) = (30)(0.5√3) = -15√3 Newton (negative because it has same direction with -y axis)

[irp]

3. F₁ = 2 N, F₂ = 4 N, F₃ = 6 N. Find both the x and y components of F₁, F₂ and F₃!

Solution

F_1x = F₁ cos 60^o = (2)(cos 60^o) = (2)(0.5) = 1 Newton (positive because it has the same direction with x axis)

F_2x = F₂ cos 30^o = (4)(cos 30^o) = (4)(0.5√3) = -2√3 Newton (negative because it has the same direction with -x axis)

F_3x = F₃ cos 60^o = (6)(cos 60^o) = (6)(0.5) = 3 Newton (positive because it has the same direction with x axis)

F_1y = F₁ sin 60^o = (2)(sin 60^o) = (2)(0.5√3) = √3 Newton (positive because it has the same direction with y axis)

F_2y = F₂ sin 30^o = (4)(sin 30^o) = (4)(0.5) = 2 Newton (positive because it has the same direction with y axis)

F_3y = F₃ sin 60^o = (6)(sin 60^o) = (6)(0.5√3) = -3√3 Newton (negative because it has the same direction with -y axis)

[irp]

[wpdm_package id=’542′]

[wpdm_package id=’554′]

1. Determine the resultant of in a line vector
2. Determine vector components
3. Determine resultant of two vectors using Pythagorean theorem
4. Determine resultant of two vectors using cosines equation
5. Determine resultant of two vectors using components of vectors

Read more

Solved problems in vectors – determine resultant of in a line vector

1. A student walks north as far as 10 meters and then to the south as far as 4 meters. Displacement of the student is…

Solution

R = 10 m – 4 m = 6 meters

Magnitude of displacement is 6 meters, direction of displacement is north.

2. F₁ = 10 N, F₂ = 15 N. Determine resultant vector…

Solution

R = 10 N + 15 N = 25 Newton

Magnitude of resultant vector is 25 Newton, direction of resultant vector is east or rightward.

[irp]

3. F₁ = 4 N, F₂ = 8 N. Determine resultant vector…

Solution

R = 8 N – 4 N = 4 Newton

Magnitude of resultant vector is 4 Newton, direction of resultant vector is east or rightward.

4. F₁ = 10, F₂ = 15 N, F₃ = 5 N. Determine resultant vector…

Solution

R = 10 N + 5 N – 15 N = 0

The magnitude of the resultant vector is 0.

[irp]

[wpdm_package id=’542′]

[wpdm_package id=’554′]

1. Determine the resultant of in a line vector
2. Determine vector components
3. Determine the resultant of two vectors using the Pythagorean theorem
4. Determine the resultant of two vectors using cosines equation
5. Determine the resultant of two vectors using components of vectors

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1. A kicked football leaves the ground at an angle θ = 30^o to the horizontal with an initial speed of 14 m/s. Calculate the final velocity before the ball hits the ground.

Known :

Angle (θ) = 30^o

Initial velocity (v_o) = 14 m/s

Acceleration of gravity (g) = 10 m/s²

Wanted : Final velocity before the ball hits the ground

Solution :

Horizontal component of initial velocity :

v_ox = v_o cos θ = (14 m/s)(cos 30^o) = (14 m/s)(0.5√3) = 7√3 m/s

Vertical component of initial velocity :

v_oy = v_o sin θ = (14 m/s)(sin 30^o) = (14 m/s)(0.5) = 7 m/s

Final velocity at vertical direction

Choose upward direction as positive and downward direction as negative.

Known :

Initial velocity (v_o) = 7 m/s (positive upward)

Acceleration of gravity (g) = –10 m/s² (negative downward)

Height (h) = 0 (object back to initial position)

Wanted : Final velocity (v_t)

Solution :

v_t² = v_o² + 2 g h = 7² + 2(-10)(0) = 49 – 0 = 49

v_t = √49 = 7 m/s

Final velocity at horizontal direction

Initial velocity at horizontal direction is 7√3 m/s. Velocity is constant so that final velocity is same as initial velocity.

Final velocity before the object hits the ground

2. A body is projected upward at an angle of 30^o with the horizontal from a building 5 meter high. Its initial speed is 10 m/s. Calculate final velocity before the object hits the ground! Acceleration of gravity is 10 m/s².

Known :

Angle (θ) = 30^o

Initial height (h_o) = 5 meters

Initial velocity (v_o) = 10 m/s

Acceleration of gravity (g) = 10 m/s²

Wanted : Final velocity

Solution :

Horizontal component of initial velocity :

v_ox = v_o cos θ = (10 m/s)(cos 30^o) = (10 m/s)(0.5√3) = 5√3 m/s

Vertical component of initial velocity :

v_oy = v_o sin θ = (10 m/s)(sin 30^o) = (10 m/s)(0.5) = 5 m/s

Final velocity at vertical direction

Known :

Initial velocity (v_o) = 5 m/s (positive upward)

Acceleration of gravity (g) = –10 m/s² (negative downward)

Height (h) = -5 m (negative because the ground is below the initial height)

Wanted : Final velocity (v_t)

Solution :

v_t² = v_o² + 2 g h = 5² + 2(-10)(-5) = 25 + 100 = 125

v_t = √125 m/s

Final velocity at horizontal direction

Final velocity at horizontal direction is 5√3 m/s.

Final velocity

3. A small ball projected horizontally with initial velocity v_o = 8 m/s from a building 12 meters high. Calculate final velocity before ball hits ground! Acceleration of gravity is 10 m/s²

Known :

Height (h) = 12 meters

Initial velocity (v_o) = 8 m/s

Acceleration of gravity (g) = 10 m/s²

Wanted : Final velocity (v_t)

Solution :

Horizontal component of initial velocity :

v_ox = v_o = 8 m/s

Vertical component of initial velocity :

v_oy = 0 m/s

Final velocity at vertical direction

calculated using equation of free fall motion.

Known :

Acceleration of gravity (g) = 10 m/s²

Height (h) = 12 m

Wanted : Final velocity (v_t)

Solution :

v_t² = 2 g h = 2(10)(12) = 240

v_t = √240 m/s

Final velocity at horizontal direction

Initial velocity at the horizontal direction is 8 m/s. Velocity is constant so that the initial velocity equals the final velocity. So final velocity at horizontal direction is 8 m/s.

Final velocity

[wpdm_package id=’534′]

[wpdm_package id=’536′]

1. Resolve initial velocity into horizontal and vertical components
2. Determine the horizontal displacement
3. Determine the maximum height
4. Determine the time interval
5. Determine the position of the object
6. Determine the final velocity

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Solved problems in projectile motion – determine the position of an object

1. A body is projected upward at an angle of 60^o to the horizontal with an initial speed of 12 m/s. Determine the position of the object after moving for 1 second! Acceleration of gravity is 10 m/s².

Known :

Angle (θ) = 60^o

Initial velocity (v_o) = 12 m/s

Time interval (t) = 1 second

Acceleration of gravity (g) = 10 m/s²

Wanted : Object position after moving during 1 second

Solution :

Horizontal component of initial velocity :

v_ox = v_o cos θ = (12 m/s)(cos 60^o) = (12 m/s)(0.5) = 6 m/s

Vertical component of initial velocity :

v_oy = v_o sin θ = (12 m/s)(sin 60^o) = (12 m/s)(0.5√3) = 6√3 m/s

Object position at horizontal direction:

Known :

Horizontal component of velocity (v_x) = 6 m/s

Time interval (t) = 1 second

Wanted : horizontal range (x)

Solution :

6 meters / second means the ball moves as far as 6 meters every 1 second. The distance of the ball after moving for 1 second is 6 meters. So the position of the ball in the horizontal direction is 6 meters.

Object position at vertical direction :

Choose upward direction as positive and downward direction as negative.

Known :

Initial velocity (v_o) = 6√3 m/s (positive upward)

Time interval (t) = 1 second

Acceleration of gravity (g) = -10 m/s² (negative downward)

Wanted : height after moving during 1 second

Solution :

h = v_o t + 1/2 g t² = (6√3)(1) + 1/2 (-10)(1²) = 6√3 + (-5)(1) = 6√3 – 5 = 6(1.7) – 5 = 10.2 – 5 = 5.2 meters.

Object position after moving for 1 second :

Horizontal displacement (x) = 6 meters

Vertical displacement (y) = 5.2 meters

[irp]

2. A body is projected upward at an angle of 30^o to the horizontal from a building 20 meters high. Its initial speed is 50 m/s. Calculate the vertical displacement after the body moving for 1 second! Acceleration of gravity is 10 m/s².

Known :

Angle (θ) = 30^o

Initial height (h_o) = 20 meters

Initial velocity (v_o) = 50 m/s

Time interval (t) = 1 second

Acceleration of gravity (g) = 10 m/s²

Wanted : Height (h)

Solution :

Vertical component of initial velocity :

v_oy = v_o sin θ = (50 m/s)(sin 30^o) = (50 m/s)(0.5) = 25 m/s

Height :

Choose upward direction as positive and downward direction as negative.

Known :

Initial velocity (v_o) = 25 m/s (positive upward)

Time interval (t) = 1 second

Acceleration of gravity (g) = -10 m/s² (negative downward)

Wanted : Height (h)

Solution :

h = v_o t + 1/2 g t² = (25)(1) + 1/2 (-10)(1²) = 25 + (-5)(1) = 25 – 5 = 20 meters.

The height of the body after moving for 1 second is 20 meters above where the body is projected or 40 meters above ground.

[irp]

3. A small ball projected horizontally with initial velocity v_o = 10 m/s from a building 10 meters high. Calculate the ball displacement after moving 1 second! Acceleration of gravity is 10 m/s²

Known :

Initial height (h) = 10 meters

Initial velocity (v_o) = 10 m/s

Time interval (t) = 1 second

Acceleration of gravity (g) = 10 m/s²

Wanted: Position of the ball after moving 1 second!

Solution :

Horizontal displacement :

Known :

Horizontal component of velocity (v_x) = 10 m/s

Time interval (t) = 1 second

Wanted: Position of the object

Solution :

10 meters/second means the object move as far as 10 meters every 1 second. Displacement after moving for 1 second is 10 meters. So horizontal displacement is 10 meters.

Vertical displacement :

Calculated as the free fall motion.

Known :

Time interval (t) = 1 second

Acceleration of gravity (g) = 10 m/s²

Wanted : Height after moving during 1 second (h)

Solution :

h = 1/2 g t² = 1/2 (10)(1²) = (5)(1) = 5 meters.

After 1 second, the object falls as far as 5 meters. Height above the ground level = 10 meters – 5 meters = 5 meters.

The position of the object after moving 1 second :

Position of the object at horizontal direction (x) = 10 meters

The position of the object at vertical direction (y) = 5 meters

[irp]

[wpdm_package id=’532′]

[wpdm_package id=’536′]

1. Resolve initial velocity into horizontal and vertical components
2. Determine the horizontal displacement
3. Determine the maximum height
4. Determine the time interval
5. Determine the position of the object
6. Determine the final velocity

Read more

Solved problems in projectile motion – determine the time interval

1. A kicked football leaves the ground at an angle θ = 30^o to the horizontal with an initial speed of 10 m/s. Calculate the time interval to reach the maximum height! Acceleration of gravity is 10 m/s².

Known :

Angle (θ) = 30^o

Initial velocity (v_o) = 10 m/s

Acceleration of gravity (g) = 10 m/s²

Wanted : Time interval to reach the maximum height

Solution :

Vertical component of initial velocity :

v_oy = v_o sin θ = (10 m/s)(sin 30^o) = (10 m/s)(0.5) = 5 m/s

Time interval to reach maximum height is determined by the vertical motion equation. Choose upward direction as positive and downward direction as negative.

Known :

Initial velocity (v_o) = 5 m/s (positive upward)

Acceleration of gravity (g) = –10 m/s² (negative downward)

Final velocity at maximum height (v_t) = 0

Wanted : time interval (t)

Solution :

v_t = v_o + g t

0 = 5 + (-10)t

0 = 5 – 10 t

5 = 10 t

t = 5/10 = 0.5 s

[irp]

2. A body is projected upward at angle of 30^o to the horizontal with an initial speed of 30 m/s. Calculate time of flight! Acceleration of gravity is 10 m/s².

Known :

Angle (θ) = 30^o

Initial velocity (v_o) = 8 m/s

Acceleration of gravity (g) = 10 m/s²

Wanted : Time interval before body hits the ground

Solution :

Vertical component of initial velocity :

v_oy = v_o sin θ = (8 m/s)(sin 30^o) = (8 m/s)(0.5) = 4 m/s

We first calculate time interval to reach the maximum height using equation of vertical motion.

Choose upward direction as positive and downward direction as negative.

Known :

Initial velocity (v_o) = 4 m/s (positive upward)

Acceleration of gravity (g) = –10 m/s² (negative downward)

Final velocity at the maximum height (v_t) = 0

Wanted : Time interval (t)

Solution :

v_t = v_o + g t

0 = 4 + (-10)t

0 = 4 – 10 t

4 = 10 t

t = 4/10 = 0,4 s

Time interval to reach the maximum height is 0.4 s.

Time in air is 2 x 0.4 s = 0.8 s.

[irp]

3. A body is projected upward at an angle of 30^o with the horizontal from a building 10 meters high. Its initial speed is 40 m/s. How long does it take the body to reach the ground? Acceleration of gravity is 10 m/s².

Known :

Angle (θ) = 30^o

Initial height (h_o) = 10 meters

Initial velocity (v_o) = 40 m/s

Acceleration of gravity (g) = 10 m/s²

Wanted : Time in air (t)

Solution :

Vertical component of initial velocity :

v_oy = v_o sin θ = (40 m/s)(sin 30^o) = (40 m/s)(0.5) = 20 m/s

We first calculate time interval to reach the maximum height using equation of vertical motion.

Choose upward direction as positive and downward direction as negative.

Known :

Initial velocity (v_o) = 20 m/s (positive upward)

Acceleration of gravity (g) = –10 m/s² (negative downward)

Final velocity at peak (v_t) = 0

Wanted : Time interval (t)

Solution :

v_t = v_o + g t

0 = 20 + (-10)t

0 = 20 – 10 t

20 = 10 t

t = 20/10 = 2 seconds

Time in air = 2 x 2 seconds = 4 seconds.

The object is 10 meters above the ground. 4 seconds is the time interval to reach a place that parallels to the initial position. The ball is still moving downward.

The time interval to reach the ground is calculated using the equation of free fall motion. 

Known :

Acceleration of gravity (g) = 10 m/s²

High (h) = 10 meters

Wanted : Time interval (t)

Solution :

h = 1/2 g t²

10 = 1/2 (10) t²

10 = 5 t²

t² = 10/5 = 2

t = √2 = 1.4 seconds

Time interval = 1.4 seconds.

Total time interval = 4 seconds + 1.4 seconds = 5.4 seconds.

[irp]

4. A small ball projected horizontally with initial velocity v_o = 15 m/s from a building 5 meters high. Calculate time in the air! Acceleration of gravity is 10 m/s²

Known :

High (h) = 5 meters

Initial velocity (v_o) = 15 m/s

Acceleration of gravity (g) = 10 m/s²

Wanted: Time in the air (t)

Solution :

Time in the air is calculated using the equation of freely falling motion.

Known :

High (h) = 5 meters

Acceleration of gravity (g) = 10 m/s²

Wanted : Time interval (t)

Solution :

h = 1/2 g t²

5 = 1/2 (10) t²

5 = 5 t²

t² = 5/5 = 1

t = √1 = 1 second

[irp]

[wpdm_package id=’531′]

[wpdm_package id=’536′]

1. Resolve initial velocity into horizontal and vertical components
2. Determine the horizontal displacement
3. Determine the maximum height
4. Determine the time interval
5. Determine the position of object
6. Determine the final velocity

Read more

Solved problems in projectile motion – determine the maximum height

1. A kicked football leaves the ground at an angle θ = 60^o with the horizontal has an initial speed of 10 m/s. Calculate the maximum height! Acceleration of gravity is 10 m/s².

Known :

Angle (θ) = 60^o

Initial speed (v_o) = 10 m/s

Wanted : Maximum height (h)

Solution :

Vertical component of initial velocity :

sin 60^o = v_oy / v_o

v_oy = v_o sin 60^o = (10)(sin 60^o) = (10)(0.5√3) = 5√3 m/s

Choose upward direction as positive and downward direction as negative.

Known :

Acceleration of gravity (g) = -10 m/s² (negative downward)

Vertical component of initial velocity (v_oy) = +5√3 m/s (positive upward)

Final velocity at the maximum height (v_ty) = 0

Wanted : Maximum height (h)

Solution :

v_t² = v_o² + 2 g h

0² = (5√3)² + 2 (-10) h

0 = 25(3) – 20 h

0 = 75 – 20 h

75 = 20 h

h = 75 / 20

h = 3.75 meter

The maximum height is 3.75 meter.

[irp]

2. A body is projected upward at angle of 30^o with the horizontal from a building 20 meter high. It’s initial speed is 4 m/s. Calculate the maximum height! Acceleration of gravity is 10 m/s².

Known :

Angle (θ) = 30^o

Initial height (h) = 20 meter

Initial velocity (v_o) = 4 m/s

Acceleration of gravity (g) = 10 m/s²

Wanted : The maximum height (h)

Solution :

Vertical component of initial velocity :

sin 30^o = v_oy / v_o

v_oy = v_o sin 30^o = (4)(sin 30^o) = (4)(0.5) = 2 m/s

Choose upward direction as positive and downward direction as negative.

Known :

Acceleration of gravity (g) = -10 m/s² (negative downward)

Vertical component of initial velocity (v_oy) = +2 m/s (positive upward)

Final velocity at maximum height (v_ty) = 0

Wanted : The maximum height

Solution :

The maximum height :

v_t² = v_o² + 2 g h

0² = 2² + 2 (-10) h

0 = 4 – 20 h

4 = 20 h

h = 4 / 20

h = 0.2 meter

The maximum height is 0.2 meter + 20 meter = 20.2 meter.

[irp]

[wpdm_package id=’528′]

[wpdm_package id=’536′]

1. Resolve initial velocity into horizontal and vertical components
2. Determine the horizontal displacement
3. Determine the maximum height
4. Determine the time interval
5. Determine the position of object
6. Determine the final velocity

Read more

Solved problems in projectile motion – determine the horizontal displacement

1. A kicked football leaves the ground at an angle θ = 60^o with the horizontal has an initial speed of 16 m/s. How long will it be before the ball hits the ground?

Known :

Angle (θ) = 60^o

Initial speed (v_o) = 16 m/s

Acceleration of gravity (g) = 10 m/s²

Wanted : Horizontal displacement (x)

Solution :

Horizontal component of initial velocity :

v_ox = v_o cos θ = (16 m/s)(cos 60^o) = (16 m/s)(0.5) = 8 m/s

Vertical component of initial velocity :

v_oy = v_o sin θ = (16 m/s)(sin 60^o) = (16 m/s)(0.5√3) = 8√3 m/s

Projectile motion could be understood by analyzing the horizontal and vertical component of the motion separately. The x motion occurs at constant velocity and the y motion occurs at constant acceleration of gravity.

Time in the air

The time it stays in the air is determined by the y motion. We first find the time using the y motion and then use this time value in the x equations (constant velocity equation).

Choose upward direction as positive and downward direction as negative.

Known :

Initial velocity (v_o) = 8√3 m/s (v_o upward)

Acceleration of gravity (g) = -10 m/s² (g downward)

Height (h) = 0 (ball is back to the same position)

Wanted : Time in air

Solution :

h = v_o t + 1/2 g t²

0 = (8√3) t + 1/2 (-10) t²

0 = 8√3 t – 5 t²

8√3 t = 5 t²

8 (1.7) = 5 t

14 = 5 t

t = 14 / 5 = 2.8 seconds

Horizontal displacement

Known :

Velocity (v) = 8 m/s

Time interval (t) = 2.8 seconds

Wanted : Displacement

Solution :

x = v t = (8 m/s)(2.8 s) = 22.4 meters

Horizontal displacement is 22.4 meters.

[irp]

2. A body is projected upward at angle of 60^o with the horizontal from a building 50 meter high. It’s initial speed is 30 m/s. Calculate the horizontal displacement! Acceleration of gravity is 10 m/s².

Known :

Angle (θ) = 60^o

High (h) = 15 m

Initial speed (v_o) = 30 m/s

Acceleration of gravity (g) = 10 m/s²

Wanted : x

Solution :

Horizontal component of initial velocity ::

v_ox = v_o cos θ = (30 m/s)(cos 60^o) = (30 m/s)(0.5) = 15 m/s

Vertical component of initial velocity :

v_oy = v_o sin θ = (30 m/s)(sin 60^o) = (30 m/s)(0.5√3) = 15√3 m/s

Time in the air

We first find the time using the y motion and then use this time value in the x equations (constant velocity equation). Choose upward as positive and downward as negative.

Known :

Initial velocity (v_o) = 15√3 m/s (positive upward)

Acceleration of gravity (g) = -10 m/s² (negative downward)

High (h) = -50 (Ground 50 meter below the initial position)

Wanted : t

Solution :

h = v_o t + 1/2 g t²

-50 = (15√3) t + 1/2 (-10) t²

-50 = 15√3 t – 5 t²

5 t² – 15√3 t – 50 = 0

Calculate time using this formula :

a = 5, b = –15√3, c = –50

Time in the air is 6.7 seconds.

Horizontal displacement :

Known :

Velocity (v) = 15 m/s

Time interval (t) = 6.7 seconds

Wanted : displacement

Solution :

s = v t = (15 m/s)(6.7 s) = 100.5 meters

Horizontal displacement is 100.5 meters.

[irp]

3. A small ball projected horizontally with initial velocity v_o = 10 m/s from a building 10 meter high. Calculate the horizontal displacement! Acceleration of gravity is 10 m/s²

Known :

High (h) = 10 m

Initial velocity (v_o) = 10 m/s

Acceleration of gravity (g) = 10 m/s²

Wanted : x

Solution :

Horizontal component of initial velocity = initial velocity = 10 m/s.

Time in the air

Time in air calculated using free fall motion equation.

Known :

Acceleration of gravity (g) = 10 m/s²

High (h) = 10 meter

Wanted : t

Solution :

h = 1/2 g t²

10 = 1/2 (10) t²

10 = 5 t²

t² = 10 / 5 = 2

t = √2 = 1.4 seconds

Horizontal displacement

Horizontal displacement calculated using equation of motion at constant velocity.

Known :

Velocity (v) = 10 m/s

Time interval (t) = 1.4 seconds

Wanted : x

Solution :

s = v t = (10 m/s)(1.4 s) = 14 meters

Horizontal displacement is 14 meters.

[irp]

[wpdm_package id=’526′]

[wpdm_package id=’536′]

1. Resolve initial velocity into horizontal and vertical components
2. Determine the horizontal displacement
3. Determine the maximum height
4. Determine the time interval
5. Determine the position of object
6. Determine the final velocity

Read more

Solved problems in projectile motion – resolve initial velocity into horizontal and vertical components

1. A kicked football leaves the ground at an angle θ = 60^o with a velocity of 10 m/s. Calculate the initial velocity components!
Known :
Angle (θ) = 60^o
Initial velocity (v_o) = 10 m/s
Wanted : v_ox dan v_oy
Solution :
Resolve the initial velocity into x component (horizontal) and y component (vertical).
sin θ = v_oy / v_o —–> v_oy = v_o sin θ
cos θ = v_ox / v_o —–> v_ox = v_o cos θ

x component (horizontal) :
v_ox = v_o cos θ = (10 m/s)(cos 60^o) = (10 m/s)(0.5) = 5 m/s

y component (vertical) :
v_oy = v_o sin θ = (10 m/s)(sin 60^o) = (10 m/s)(0.5√3) = 5√3 m/s

[irp]

2. An object leaves ground at an angle θ = 30^o with y component of the velocity 10 m/s. Calculate initial velocity !
Known :
Angle (θ) = 30^o
y component (v_oy) = 10 m/s
Wanted : Initial velocity (v_o)
Solution :
v_oy = v_o sin θ
10 = (v_o)(sin 30^o)
10 = (v_o)(0.5)
v_o = 10 / 0.5
v_o = 20 m/s

[irp]

3. Horizontal component of initial velocity is 30 m/s and vertical component of initial velocity is 40 m/s. Calculate initial velocity.
Known :
Horizontal component of initial velocity (v_ox) = 30 m/s
Vertical component of initial velocity (v_oy) = 40 m/s
Wanted : Initial velocity (v_o)
Solution :
v_o² = v_ox² + v_oy² = 30² + 40² = 900 + 1600 = 2500
v_o = √2500
v_o = 50 m/s

[irp]

4. A small ball projected horizontally with initial velocity v_o = 6 m/s. Calculate x component and y component of initial velocity.
Known :
Initial velocity (v_o) = 6 m/s
Wanted : vox and v_oy
Solution :
Ball move horizontally so that horizontal component of velocity (v_ox) = initial velocity (v_o) = 6 m/s. Vertical component of velocity (v_oy) = 0.

[irp]

[wpdm_package id=’545′]

[wpdm_package id=’536′]

1. Resolve initial velocity into horizontal and vertical components
2. Determine the horizontal displacement
3. Determine the maximum height
4. Determine the time interval
5. Determine the position of object
6. Determine the final velocity

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