Question

The viewing room window is 1.70 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 13.8 m/s.

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

Answer 1

The acceleration due to gravity near the earths surface: $g=9.81\ m/s^2$  

Ignoring air resistance

$$a=-g=dv/dt$$

Then

$$dv=adt=-gdt$$

$$v(t)=\int{dv}=-\int{g}dt=v_0-gt=dh/dt$$

$$h(t)=\int{dh}=\int{v}dt=\int{v_0-gt}dt=h_0+v_0t-{gt^2 \over 2}$$

Given that $h_0=0, h(t_1)=1.70\ m, v(t_1)=-13.8\ m/s$

We have the system

$$1.70\ m=0\ m+v_0t_1-{9.81\ m/s^2\cdot t_1^2 \over 2}$$

$$-13.8\ m/s=v_0-9.81\ m/s^2\cdot t_1$$

$$t_1={v_0+13.8 \over 9.81}$$

$${9.81({v_0+13.8})^2 \over 2(9.81)^2}-{v_0(v_0+13.8) \over 9.81}+1.7=0, v_0>0$$

$$v_0^2+27.6v_0+190.44-2v_0^2-27.6v_0+33.354=0$$

$$v_0^2=223.794$$

$$v_0=\pm \sqrt{223.794}$$

Since $v_0>0,$ we take

$$v_0=\sqrt{223.794}\ m/s\approx14.96\ m/s$$