A tennis ball on Mars, where the acclertation due to gravity is 0.379g, and air resistance is negligible, is hit directly upwards and returns to the same level at 8.5 seconds time.
A) how high above its original point did the ball go?
B) what was it initial velocity after it was hit?

Question

A tennis ball on Mars, where the acclertation due to gravity is 0.379g, and air resistance is negligible, is hit directly upwards and returns to the same level at 8.5 seconds time.
A) how high above its original point did the ball go?
B) what was it initial velocity after it was  hit?

anonymous · Answer

MY variables are 
a= 3.72 ms^2
T=8.5 seconds
Vf= 0

I converted acceleration to m/s^2 and got 3.72 m/s^2
and i assumed Vyf is =0

anonymous · Answer

@sammixboo do you have time to take a look?

anonymous · Answer

using the kinematic equation d=vit+1/2at^2 i got
33.6 as my distance

anonymous · Answer

@ganeshie8 am i attacking this problem the right way?

ganeshie8 · Answer

8.5 seconds is the total time for going up and falling all the way back again
so you need to take t = 8.5/2 for finding the maximum height

anonymous · Answer

to get my distance, started from the top, which gave me my variables
vi=0
a=3.72
t=4.25

i use $$\delta d= 67.13$$ and $$\delta t = 8.5$$

i use the equation d= vit+1/2at^2 but i get a vi of 67.13 ms
which doesnt seem right

ganeshie8 · Answer

you may start by finding the initial velocity

ganeshie8 · Answer

$v=u+at  $
plugin  $v = 0$, $a = -3.72$ and  $t = 4.25$

ganeshie8 · Answer

$0 = u - 3.72 (4.25)$

$u = ?$

anonymous · Answer

A) 33.6
b)15.81

ganeshie8 · Answer

Looks good!

anonymous · Answer

thanks, it seems correct now!

ganeshie8 · Answer

yw