Posted by Wes on .
A rock is dropped from a treetop 19.6 m high, and then,
1.00 s later, a second rock is thrown down. With what
initial velocity must the second rock be thrown if it is to
reach the ground at the same time as the first?
I'm not sure how to find the initial velocity of the thrown rock, since it's not 0
What I have so far:
1st rock:
Yo=o
Yf=19.6 m
Vo=o
Vf=
T=
A= 9.80 m/s^2
I think I've found the final velocity and time of the 1st rock
(Vf^2=Vo^22g(YYo)
or final velocity squared = initial velocity(0 m/s)^2 2(9.80)(19.60)
which is = to 19.6 m/s
Time = 2 seconds
(Vf=Vogt, Vf/Vog=T, 19.6 m/s /9.80 m/s^2, 2 seconds
How do I find the initial velocity of a throw object when I know:
Yo=o
Yf=19.6 m
T= 21 = 1 second
A /9.80 m/s^2
Thanks!

physics 
Denn,
Wess, there are 2 cases in this prob:
Case 1
U nid to find how long the first rock took to hit the ground: use the formular s=ut+0.5gt^2 where s=19.6m, u=0, t=?
I gor t=1.97s
Case 2
The second rock took 1.97s1s to hit the grnd, use same formula & find u 
physics 
Wes,
Thank you very much!