Basic Physics

# 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

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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.

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.

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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

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