Ed jumped 69.5 m and was airborne for 3.66 sec. What was the maximum height he reached?

Question

If you coudl show your work it'd be greatly appreciated!! =)

Answer 1

I did?
Was it on the Moon or here on Earth?

I assume it was Earth g=9.81m/s^2

If I was in the air for 3.66 sec total I probably reached the top at 3.66/2=1.83s (assuming the same elevation )

h=0.5 g t^2
h=0.5 x 9.81(1.83s)^2=
h=16.4m

Answer 2

I think we, your answerers, would greatly appreciate it if you copied the question from your textbook word for word. There is a large set of things missing from this question. They have to be provided before it can be answered.

Here are some of them:

What was the initial vertical velocity uv? We need that to figure KE(v) = 1/2 m(uv)^2, which is the initial vertically derived kinetic energy of the jumper at the start. The initial horizontal velocity is uh = U cos(theta) = S/t; where S = 69.5 m and t = 3.66 sec. But we have no way to figure uv.

Is force of gravity the only force acting on Ed? What about drag friction (D = 1/2 rho Cd A V^2); where rho is air/gas density, Cd is drag coefficient, A is cross sectional area, and V is the time dependent velocity over Ed's path during the jump. V slows down a bit as drag forces decelerate the jump. This in turn reduces the drag forces....it's cyclical. Drag also acts in the vertical direction to offset the pull of gravity a bit.

Without drag (e.g., in a vacuum), we still have an issue with your question. Normally, in a vacuum, we could set kinetic energy KE(v) = 1/2 m(uv)^2 = mgh = PE, which is potential energy stemming from the work against gravity. In which case h = 1/2 uv^2/g and that would be the maximum height because all the KE stemming from uv would be converted to PE at that height.

But you failed to tell us where Ed was jumping from so we could find g. g = 9.81 m/sec, but if and only if the jumper is jumping from the surface of Earth or body of similar radius R and mass M. It's about 1/6 that if jumping from the surface of the Moon for example.

Suggest you add the appropriate information; so we can answer it properly.