Derive a general equation for the time of flight of the banana en route to the monkey in terms of only the velocity, distance, and height (as labelled in the graph above).
hint: if you are doing everything right, there should be a lot of cancellations resulting in a relatively simple solution
I do not do magic. I do not have your diagram. Is the monkey falling at the same time you release the banana? If so, then shoot straight at the monkey because the banana and the monkey fall at the same rate. Thus you only need the time for the monkey to fall H
h = (1/2) g t^2
t = sqrt (2 h/g)
then
theta = tan^-1 (h/x) where x is distance along ground to monkey tree
and
u = V cos theta = x/t
so
V cos theta = x/t = x/sqrt(2 h/g)
