Determine time interval of projectile motion

Solved problems in projectile motiondetermine the time interval

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 :

Solving projectile motion problems – determine time interval 1Vertical 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 :

Solving projectile motion problems – determine time interval 2Vertical 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.

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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 :

Solving projectile motion problems – determine time interval 3Time 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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  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