Help with Physics?

Question

In a tennis match, a player wins a point by hitting the ball sharply to the ground on the opponent's side of the net.

If the ball bounces upward from the ground with a speed of 18 m/s and is caught by a fan in the stands with a speed of 12 m/s, how high above the court is the fan? Ignore air resistance.

Help please. I just can't figure this one out.

Answer 1

u = 18m/s , v = 12m/s

v² = u² - 2gh
so
h = (u² - v²)/2g ≈ 9.2m

,.,.,.,.,.,.,.,

Answer 2

u = 18 m/s
v = 12m/s

v^2 - u^2 = 2as
12^2 - 18^2 = 2(-9.8)s
s = 18^2 - 12^2 / 19.6
s = 9.183 m

Answer 3

SinceV=gt and V=v1-v2 we have the time in flight
t=(v1-v2)/g
and we know that the hieght for fre fall is
h=(1/2) gt^2
then
h=(1/2)g((v1-v2)/g)^2
h=((v1-v2))^2 / (2g)

Thumbs up to Black Wolf!

However h=1.83 m
(unless the acceleration due to gravity has become 1.96 m/s^2)

Answer 4

Try it the potential energy way, . This has to be the potential energy at height h, converted to energy if dropped from h
v^2=2gh=18^2=2*9.8*h::::h=:::::192/9.8=19.6m
v^2=12^2=2*9.8*f::::::::f=72/9.8=7.3m/s
19.6-72/9.8=19.6-7.3=12.3m answer