2 possible initial velocities

[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.

 

[baba89]

I would first clarify the question and gather more information. Are we assuming that the arrow is shot straight up? If so, then the initial velocity would be the same as the final velocity at the top of the trajectory, which is 0 m/s. This would mean that the arrow would take 3 seconds to reach its maximum height of 35m and another 3 seconds to return to the same height, for a total time of 6 seconds.

However, if the arrow is shot at an angle, then the initial velocity would be different and the trajectory would be more complex. In this case, we would need to know the angle at which the arrow was shot and the initial velocity in order to accurately calculate the time it takes for the arrow to return to the same height.

In summary, the initial velocity will depend on the angle at which the arrow is shot and cannot be determined without additional information.