Question

A tennis player throws a ball vertically upwards at 4.90 m/s. How long, in seconds, does the ball take to return to his hand (assuming it has not moved since releasing the ball)?
Assume a gravitational acceleration of g = 9.8 m/s^2 and give your answer to 2 decimal places.

Answer 1

If ball return to the hand of player then we can say that displacement of ball will be zero

so here we will have

$\Delta y = v t + \frac{1}{2} at^2$

so here we will have

$0 = 4.90(t ) + \frac{1}{2}(-9.8)t^2$

$0 = 4.90- 4.90 t$

$t = 1.0 s$

so it will take 1 second to come back in its hand