Answer to Question #94535 in Classical Mechanics for Kevin

Question

You are thinking about taking gymnastics, so you go to the facility and get an idea of what to expect by looking out from the viewing room. The viewing room window is 3.50 m above the trampoline directly below, so it is perfect for viewing the the facility. Occasionally someone jumps past the window and then back down. On one occurrence a gymnast went up past the window and came back down; as he passed the window on the way down, you notice that his speed is 12.6 m/s.

What must have been his initial speed coming off the trampoline?

Answer 1

Using energy conservation principle:

$$E_{init}=E_{final}$$ $$E_{kin_0}=E_{kin}+E_{pot}$$ $$\frac{mv_0^2}{2}=\frac{mv^2}{2}+mgh$$ $$v_0=\sqrt{v^2+2gh}$$

$$v_0=\sqrt{12.6^2+2*9.8*3.5}=15 m/s$$

Answer: Initial speed: 15 m/s