Determine the maximum height of projectile motion

Solved problems in projectile motiondetermine the maximum height

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 :

Solving projectile motion problems – determine the maximum height 1Vertical 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

Initial velocity (vo) = 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]

[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

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Determine horizontal displacement of projectile motion

Solved problems in projectile motiondetermine the horizontal displacement

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)

Solving projectile motion problems – determine horizontal displacement 1Solution :

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 :

Solving projectile motion problems – determine horizontal displacement 2Horizontal 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

Solving projectile motion problems – determine horizontal displacement 1

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 :

Solving projectile motion problems – determine horizontal displacement 4Horizontal 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

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Resolve initial velocity into horizontal and vertical components of projectile motion

Solved problems in projectile motionresolve initial velocity into horizontal and vertical components

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 :
Solving projectile motion problems – resolve initial velocity into horizontal and vertical components 1Resolve 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]

[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

Read more

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