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/s^{2}.

__Known :__

Angle (θ) = 30^{o}

Initial velocity (v_{o}) = 10 m/s

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

__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/s^{2} (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/s^{2}.

__Known :__

Angle (θ) = 30^{o}

Initial velocity (v_{o}) = 8 m/s

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

__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/s^{2} (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/s^{2}.

__Known :__

Angle (θ) = 30^{o}

Initial height (h_{o}) = 10 meters

Initial velocity (v_{o}) = 40 m/s

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

__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/s^{2} (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/s^{2}

High (h) = 10 meters

__Wanted :__ Time interval (t)

__Solution :__

h = 1/2 g t^{2}

10 = 1/2 (10) t^{2}

10 = 5 t^{2}

t^{2 }= 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/s^{2}

__Known :__

High (h) = 5 meters

Initial velocity (v_{o}) = 15 m/s

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

__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/s^{2}

__Wanted :__ Time interval (t)

__Solution :__

h = 1/2 g t^{2}

5 = 1/2 (10) t^{2}

5 = 5 t^{2}

t^{2 }= 5/5 = 1

t = √1 = 1 second

[irp]

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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