Constant Acceleration-Find Initial Velocity

1. Jun 24, 2010

sona1177

1. The problem statement, all variables and given/known data
A juggler throws a ball straight up into the air with a speed of 10 m/s. With what speed would she need to throw a second ball half a second later, starting from the same position as the first, in order to hit the first ball at the top of the projectory?

My problem is I am getting different answers when I use different equations all involving time.

Given/Known
First Ball:

(Vy)i=10 m/s
(Vy)f= 0 m/s
ay=-9.8 m/s2

Second Ball
(Vy)f=0 m/s
ay=-9.8 m/s2

2. Relevant equations
V = Vo + at
Y - Yo = Vot + .5at2
v2 = vo2 + 2a(Y - Yo)
Y - Yo = .5(Vo + V)t

3. The attempt at a solution

I found the time it takes for the first ball to reach the top of the projectory using equation (V=Vo + at) where V= 0, Vo=10, and a=-9.8. The value I got was 1.02 seconds. I then thought that since the second ball is thrown half a second later, the time for it to reach and hit the top of the projectory should be around .5 seconds. So then I used the same equation: V=Vo + at where V=0, Vo=unknown, a=-9.8, and t=.5 seconds. The value I got was 4.9 m/s. This is wrong. But when I solved for the displacement for the first ball (from initial point to top of projectory) using equation v2 = vo2 + 2a(Y - Yo) where V= 0, Vo=unknown, a=-9.8 the value I got for the displacement was 5.1 m. Now when I plug this value to find the initial velocity of the second ball using equation Y - Yo = Vot + .5at2 where Y-Yo=5.1 m, Vo=unknown, t=.5 seconds (because the second ball is thrown half a second later than the first so since the first one takes around one second the second ball should take around .5 seconds thrown from the same position to hit the first ball), and a=
-9.8 the value for initial velocity becomes 12.64 m/s. Why am I getting two different answers using two different equations? Shouldn't both give me the same value of time? This is very frustrating.

2. Jun 24, 2010

rl.bhat

When you use V = Vo + at, final velocity becomes zero a different height than the first one. so they won't meet. They will meet only when their displacement is the same in the given time interval.

3. Jun 24, 2010

sona1177

I feel so dumb right now! :(
Thank you so much! I should have asked you 2 hours ago when I first started struggling. And yes, I checked, the answer is around 12 m/s, so this can only be true when you take into account displacement (the second equation I used). Thank you for reminding me that If you use just velocities this is wrong because the heights are different. You are a savior.