How Do You Calculate the Time for an Arrow to Return to the Same Height?

[krausr79]

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?

 

[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.