A tennis ball was tossed from a height of 1.15 m with an initial velocity of 16.

Question

A tennis ball was tossed from a height of 1.15 m with an initial velocity of 16.0 m/s at an angle 31.0° above the horizontal. The ball hits the ground at a horizontal distance of 26.8 m from the launch point. Use 10.0 N/kg for g.

a) Find in m/s the magnitude of the velocity of the ball when it is at the top most position of its trajectory.

b) Find in m/s2 the magnitude of the acceleration of the ball when it is at the top most position of its trajectory.

c) How long in seconds is the ball in flight?

d) What is in m/s the magnitude of the velocity with which the ball hits the ground?

e) What is in m the highest point relative to ground reached by the ball?

Explanation / Answer

Here ,

height , h = 1.15 m

intial speed , u = 16 m/s

theta = 31 degree

a)

at the top most postion ,

horizontal velocity = 16 * cos(31)

horizontal velocity = 13.7 m/s

b)

as the acceleration is constant throughout the motion ,

accelration , a = 9.8 m/s^2 downwards

c) time of flight = distance/speed

time of flight = 26.8/(16 * cos(31))

time of flight = 1.95 s
d)

let the final velocity of ball is v

v^2 = 16^2 + 2 * 9.8 * 1.15

v = 16.7 m/s

the magnitude of the final velocity is 16.7 m/s

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