# Determine the maximum height of projectile motion

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/s^{2}.

__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/s^{2 }*(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_{t}^{2} = v_{o}^{2} + 2 g h

0^{2} = (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.

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/s^{2}.

__Known :__

Angle (θ) = 30^{o}

Initial height (h) = 20 meter

Initial velocity (v_{o}) = 4 m/s

Acceleration of gravity (g) = 10 m/s^{2}

__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/s^{2 }*(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_{t}^{2} = v_{o}^{2} + 2 g h

0^{2} = 2^{2} + 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.

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- Resolve initial velocity into horizontal and vertical components
- Determine the horizontal displacement
- Determine the maximum height
- Determine the time interval
- Determine the position of object
- Determine the final velocity