Question

a) a driver of mass 55kg needs to jump to a height of 1.1m above her take off height, at which point her speed needs to be 0.5m/s^. How fast just she be travelling the instant her feet leave the diving board?

So we know eg=mgh =9.8*55*1.1

I assumed we're trying to find the power

So v=d/t and then we have 0.5=101/t. So then we find t and divide it by the eg? I'm not sure if that's correct

b) at what speed does she land in the water?

I'm bit sure how to do this, I assume you use the p from question 2 to find the time. And then use the distance formula to find v?

Answer 1

energy at top=takeoffienergy
mgh+1/2 m (j0.5^2)=1/2 m (vi)^2
solve for vi. Notice mass m divides out.
now speed at the water...how much lower is the water from the diving board. If you assume it is at the
KEwater=initialKE + initial PE
1/2 vf^2=1/2 m vi^2+mg(ho)
where ho is the initial height of the diving board, solve for vf