# Free Fall Tennis Ball Question

1. Aug 14, 2013

### PerryKid

1. The problem statement, all variables and given/known data

A tennis ball is thrown upward with and reaches height of 18 m. What maximum height can reach this ball on the where acceleration of free fall is 6 times less than g? In both cases initial velocity is same. [sic] (excuse my physics teacher's grammar, don't shoot the messenger! :tongue:)

g= 9.81 m/s2
x= distance or displacement
v= velocity
a=acceleration (which is g)

Max Height is at V=0

2. Relevant equations

$V=\sqrt{2gx}$
$ΔX=V_it + (1/2) at^2$

3. The attempt at a solution

$9.81/6=1.64$

$V^2=2(1.64)x$

Initial velocity in a free fall is 0 m/s, right? If so, v=0 and then x=0

In such a case, it would be odd for the max height to be 0 m.

Last edited: Aug 14, 2013
2. Aug 14, 2013

### voko

If the ball is thrown upward, then its initial velocity cannot be zero.

Note the formulation does not require that the initial velocity be zero; it just says it is the same in both cases.

3. Aug 14, 2013

### PerryKid

So, I can just use any velocity?

Or must I use a velocity that intersects?

Considering time is neither given nor required, can I use

$V_f = V_i+at$?

$V_f=0$ to find the maximum.

I use the gravities as the slope. However, the lines intersect at the origin, where time and velocity equal 0.

$0 = -1.64 t$

$0 = -9.81 t$

Last edited: Aug 14, 2013
4. Aug 14, 2013

### Staff: Mentor

You can find the initial velocity using your first Relevant Equation for the stated conditions that the ball reaches a height of 18 m when gravity is g.

Note that you are taking advantage of the fact that ideal free-fall trajectories are symmetric with respect to time reversal, so if a ball launched upwards with some velocity V reaches maximum height H, then a ball dropped from height H will reach a final velocity V just before impact with the ground.

5. Aug 15, 2013

### voko

Does that really matter what the initial velocity is numerically? You know it is the same. That means the initial kinetic energy is the same.