posted by on .

I am doing the monkey and hunter question, and I need to find out what the minimum speed the bullet has to be..
The variables I have are g, Vo, theta, Dx(horizontal distance from monkey), and Dy (vertical distance from monkey).

I need to derive Vo in the form of (Dx/cos0)(sqrt(g/2Dy))...

I have managed to derive Vo = Vo = sqrt((0.5(Dx/cos0/sinO))g) from Dx = (VoCosθ)(2((Vo)(Sinθ))/g), but its not the one I need..

Thank you so much for taking the time to help me

If the shot is fired as the monkey falls, fire exactly at the monkey because the bullet and the monkey will both accelerate down at g and meet.
Dy/Dx = tan T
Dy = Dx tan T

It does not matter what Vo is as long as the bullet and monkey do not hit the ground before colliding.
So for monkey:
Dy = (1/2) g t^2
so t = sqrt(2 Dy/g)
then for bullet
h = 0 when
0 = (Vo sin Theta) t - 1/2 g t^2

same t for both
(g/2) (2 Dy/g) = Vo sin Theta sqrt (2 Dy/g)

(g/2) sqrt(2 Dy/g) = Vo sin Theta

sqrt (g Dy/2) = Vo sin Theta

Vo = [ sqrt g Dy/2 ] / sin Theta

the very smallest Vo will be as theta approaches 90 degrees (shoot straight up from under tree)

In that case Vo = sqrt (g Dy/2)