# Homework Help: 2 possible initial velocities

1. Sep 19, 2008

### krausr79

1. The problem statement, all variables and given/known data
I am given a height and a time it took to get there under 'free fall' conditions. 35m after 3sec An arrow is shot into the air and is 35m high after 3 sec. How long until it returns (falls to) to 35m high again?

2. Relevant equations
X=Xo+Vo*T+1/2*A*T^2

3. The attempt at a solution
X = 35m
Xo = 0m
V = ?
Vo = To find
A = -9.8 m/s2
t = 3 sec

35=0+3*Vo+1/2*-9.8*3^2
solved, Vo = 26.36

Then, with time as a variable:
35=0+26.36*t+1/2*-9.8*t^2
0=-35+26.36*t-4.9*t^2
Using the quadratic formula gives me 3 seconds and 2.38 seconds, but that means it assumed that 3 seconds was the return time and 2.38 seconds was when the arrow first arrived. 3 seconds was meant to be the first arrival. There must be a second initial velocity that causes 3 seconds to be the first arrival, but the equation doesn't suggest any way to find it. How is that possible? What is the real initial velocity?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 19, 2008

### bdforbes

There is only one possible initial velocity, since the equation of motion is linear in v_0. I calculated 55.77m/s where you got 26.36m/s. Check your working.

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