## Determine resultant of two vectors using Pythagorean theorem

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

[irp]

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

[irp]

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]

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## Determine vector components

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

Solution

Fx = F cos 30o = (20)(cos 30o) = (20)(0.53) = 103 Newton

Fy = F sin 30o = (20)(sin 30o) = (20)(0.5) = 10 Newton

[irp]

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

Solution

F1x = F1 cos 60o = (20)(cos 60o) = (20)(0.5) = -10 Newton (negative because it has same direction with -x axis)

F2x = F2 cos 60o = (30)(cos 60o) = (30)(0.5) = -15 Newton (negative because it has same direction with -x axis)

F1y = F1 sin 60o = (20)(sin 60o) = (20)(0.53) = 103 Newton (positive because it has same direction with y axis)

F2y = F2 sin 60o = (30)(sin 60o) = (30)(0.53) = -153 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

F1x = F1 cos 60o = (2)(cos 60o) = (2)(0.5) = 1 Newton (positive because it has the same direction with x axis)

F2x = F2 cos 30o = (4)(cos 30o) = (4)(0.53) = -23 Newton (negative because it has the same direction with -x axis)

F3x = F3 cos 60o = (6)(cos 60o) = (6)(0.5) = 3 Newton (positive because it has the same direction with x axis)

F1y = F1 sin 60o = (2)(sin 60o) = (2)(0.53) = 3 Newton (positive because it has the same direction with y axis)

F2y = F2 sin 30o = (4)(sin 30o) = (4)(0.5) = 2 Newton (positive because it has the same direction with y axis)

F3y = F3 sin 60o = (6)(sin 60o) = (6)(0.53) = -33 Newton (negative because it has the same direction with -y axis)

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## 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]

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## Determine final velocity of projectile motion

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

Known :

Angle (θ) = 30o

o) = 14 m/s

10 m/s2

Wanted : Final velocity before the ball hits the ground

Solution :

Horizontal component of initial velocity :

vox = vo cos θ = (14 m/s)(cos 30o) = (14 m/s)(0.53) = 73 m/s

Vertical component of initial velocity :

voy = vo sin θ = (14 m/s)(sin 30o) = (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 (vo) = 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 (vt)

Solution :

vt2 = vo2 + 2 g h = 72 + 2(-10)(0) = 49 – 0 = 49

vt = √49 = 7 m/s

Final velocity at horizontal direction

Initial velocity at horizontal direction is 73 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 30o 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) = 30o

Initial height (ho) = 5 meters

Initial velocity (vo) = 10 m/s

Acceleration of gravity (g) = 10 m/s2

Wanted : Final velocity

Solution :

Horizontal component of initial velocity :

vox = vo cos θ = (10 m/s)(cos 30o) = (10 m/s)(0.53) = 53 m/s

Vertical component of initial velocity :

voy = vo sin θ = (10 m/s)(sin 30o) = (10 m/s)(0.5) = 5 m/s

Final velocity at vertical direction

Known :

Initial velocity (vo) = 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 (vt)

Solution :

vt2 = vo2 + 2 g h = 52 + 2(-10)(-5) = 25 + 100 = 125

vt = √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 vo = 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 (vo) = 8 m/s

Acceleration of gravity (g) = 10 m/s2

Wanted : Final velocity (vt)

Solution :

Horizontal component of initial velocity :

vox = vo = 8 m/s

Vertical component of initial velocity :

voy = 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 (vt)

Solution :

vt2 = 2 g h = 2(10)(12) = 240

vt = √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

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## Determine the position of an object in projectile motion

1. A body is projected upward at an angle of 60o 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) = 60o

Initial velocity (vo) = 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 :

vox = vo cos θ = (12 m/s)(cos 60o) = (12 m/s)(0.5) = 6 m/s

Vertical component of initial velocity :

voy = vo sin θ = (12 m/s)(sin 60o) = (12 m/s)(0.53) = 63 m/s

Object position at horizontal direction:

Known :

Horizontal component of velocity (vx) = 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 (vo) = 63 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 = vo t + 1/2 g t2 = (63)(1) + 1/2 (-10)(12) = 63 + (-5)(1) = 63 – 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 30o 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) = 30o

Initial height (ho) = 20 meters

Initial velocity (vo) = 50 m/s

Time interval (t) = 1 second

Acceleration of gravity (g) = 10 m/s2

Wanted : Height (h)

Solution :

Vertical component of initial velocity :

voy = vo sin θ = (50 m/s)(sin 30o) = (50 m/s)(0.5) = 25 m/s

Height :

Choose upward direction as positive and downward direction as negative.

Known :

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

Time interval (t) = 1 second

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

Wanted : Height (h)

Solution :

h = vo 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 vo = 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 (vo) = 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 (vx) = 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

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## Determine time interval of projectile motion

1. A kicked football leaves the ground at an angle θ = 30o 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) = 30o

Initial velocity (vo) = 10 m/s

Acceleration of gravity (g) = 10 m/s2

Wanted : Time interval to reach the maximum height

Solution :

Vertical component of initial velocity :

voy = vo sin θ = (10 m/s)(sin 30o) = (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 (vo) = 5 m/s (positive upward)

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

Final velocity at maximum height (vt) = 0

Wanted : time interval (t)

Solution :

vt = vo + 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 30o 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) = 30o

Initial velocity (vo) = 8 m/s

Acceleration of gravity (g) = 10 m/s2

Wanted : Time interval before body hits the ground

Solution :

Vertical component of initial velocity :

voy = vo sin θ = (8 m/s)(sin 30o) = (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 (vo) = 4 m/s (positive upward)

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

Final velocity at the maximum height (vt) = 0

Wanted : Time interval (t)

Solution :

vt = vo + 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 30o 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) = 30o

Initial height (ho) = 10 meters

Initial velocity (vo) = 40 m/s

Acceleration of gravity (g) = 10 m/s2

Wanted : Time in air (t)

Solution :

Vertical component of initial velocity :

voy = vo sin θ = (40 m/s)(sin 30o) = (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 (vo) = 20 m/s (positive upward)

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

Final velocity at peak (vt) = 0

Wanted : Time interval (t)

Solution :

vt = vo + 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 vo = 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 (vo) = 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]

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## Determine the maximum height of projectile motion

1. A kicked football leaves the ground at an angle θ = 60o 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) = 60o

Initial speed (vo) = 10 m/s

Wanted : Maximum height (h)

Solution :

Vertical component of initial velocity :

sin 60o = voy / vo

voy = vo sin 60o = (10)(sin 60o) = (10)(0.53) = 53 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 (voy) = +53 m/s (positive upward)

Final velocity at the maximum height (vty) = 0

Wanted : Maximum height (h)

Solution :

vt2 = vo2 + 2 g h

02 = (53)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 30o 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) = 30o

Initial height (h) = 20 meter

o) = 4 m/s

Acceleration of gravity (g) = 10 m/s2

Wanted : The maximum height (h)

Solution :

Vertical component of initial velocity :

sin 30o = voy / vo

voy = vo sin 30o = (4)(sin 30o) = (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 (voy) = +2 m/s (positive upward)

Final velocity at maximum height (vty) = 0

Wanted : The maximum height

Solution :

The maximum height :

vt2 = vo2 + 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]

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## Determine horizontal displacement of projectile motion

1. A kicked football leaves the ground at an angle θ = 60o 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) = 60o

Initial speed (vo) = 16 m/s

Acceleration of gravity (g) = 10 m/s2

Wanted : Horizontal displacement (x)

Solution :

Horizontal component of initial velocity :

vox = vo cos θ = (16 m/s)(cos 60o) = (16 m/s)(0.5) = 8 m/s

Vertical component of initial velocity :

voy = vo sin θ = (16 m/s)(sin 60o) = (16 m/s)(0.53) = 83 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 (vo) = 83 m/s (vo 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 = vo t + 1/2 g t2

0 = (83) t + 1/2 (-10) t2

0 = 83 t – 5 t2

83 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 60o 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) = 60o

High (h) = 15 m

Initial speed (vo) = 30 m/s

Acceleration of gravity (g) = 10 m/s2

Wanted : x

Solution :

Horizontal component of initial velocity ::

vox = vo cos θ = (30 m/s)(cos 60o) = (30 m/s)(0.5) = 15 m/s

Vertical component of initial velocity :

voy = vo sin θ = (30 m/s)(sin 60o) = (30 m/s)(0.53) = 153 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 (vo) = 153 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 = vo t + 1/2 g t2

-50 = (153) t + 1/2 (-10) t2

-50 = 153 t – 5 t2

5 t2153 t – 50 = 0

Calculate time using this formula :

a = 5, b = –153, 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 vo = 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 (vo) = 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]

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## Resolve initial velocity into horizontal and vertical components of projectile motion

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

x component (horizontal) :
vox = vo cos θ = (10 m/s)(cos 60o) = (10 m/s)(0.5) = 5 m/s

y component (vertical) :
voy = vo sin θ = (10 m/s)(sin 60o) = (10 m/s)(0.5√3) = 5√3 m/s

[irp]

2. An object leaves ground at an angle θ = 30o with y component of the velocity 10 m/s. Calculate initial velocity !
Known :
Angle (θ) = 30o
y component (voy) = 10 m/s
Wanted : Initial velocity (vo)
Solution :
voy = vo sin θ
10 = (vo)(sin 30o)
10 = (vo)(0.5)
vo = 10 / 0.5
vo = 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 (vox) = 30 m/s
Vertical component of initial velocity (voy) = 40 m/s
Wanted : Initial velocity (vo)
Solution :
vo2 = vox2 + voy2 = 302 + 402 = 900 + 1600 = 2500
vo = √2500
vo = 50 m/s

[irp]

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

[irp]

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