A tennis ball is hit when it is 1.30 meters above the court. The ball's velocity leaving the racket is 8.08 m/s horizontally, how far away does the ball land?

Answer: 4.2 m

Explanation:

The time taken by the tennis ball to cover vertical displacement 1.30 m would be the same time in which ball would cover horizontal distance.

Initial velocity of the ball, u = 8.08 m/s in horizontal direction.

Initial velocity in the vertical direction is 0.

consider upward direction as positive and downward direction as negative. The ball free falls under gravity in the vertical direction.

Using the second equation of motion,

y = u t + 0.5 at²

⇒-1.30 m = 0 + 0.5 * (-9.8m/s²) t²

⇒t = 0.52 s

There is no external force acting in the horizontal direction, thus horizontal distance covered is:

s = u t ⇒ s = 8.08 m/s * 0.52 s = 4.2 m

Thus, the tennis ball lands 4.2 m far away.