- #1
krausr79
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Homework Statement
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?
Homework Equations
X=Xo+Vo*T+1/2*A*T^2
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?