# Falling monkeys! (Projectile motion problem)

1. Oct 27, 2009

### Greywolfe1982

1. The problem statement, all variables and given/known data

http://img202.imageshack.us/img202/3459/phys.png [Broken]

2. Relevant equations

$$\Delta$$Y=VoYt+1/2gt2

3. The attempt at a solution

Rearranging the equation (using $$\Delta$$Y=h), I get $$\sqrt{}2h/g$$ which, from what I understand, means it takes $$\sqrt{}2h/g$$ seconds for the monkey to reach the point at which the coconut was thrown. My problem is moving on from here. All I know is that the monkey is at the original height of the coconut and the coconut is some portion of b from it's starting point (for what it's worth, my guess is 9, but I'd rather not lose points that I don't have to trying to guess). Where should I go from here?

Thanks!

Last edited by a moderator: May 4, 2017
2. Oct 27, 2009

### rl.bhat

If the monkey catches the ball at a distance y from the ball, for monkey
Δy = h + y = .....?.......(1)
For the ball
y =.........? (2)
Solve these two equations to find t.

3. Oct 28, 2009

### Greywolfe1982

For monkey
ΔY = h + y
ΔY=1/2gt2+Vimt+1/2gt2
h is 1/2gt2, y is Vimt+1/2gt2 (Vim being the monkeys velocity after falling h meters.)

For ball
ΔY = y
ΔY = Vibt+1/2gt2

ΔY = ΔY
1/2gt2+Vimt+1/2gt2=Vibt+1/2gt2

And now I'm lost again. I tried figuring out the monkeys velocity once it reached the original height of the ball (a=Vf-Vi/t) and got
V=g$$\sqrt{}2h/g$$, put that into the monkey equation and got ΔY=g$$\sqrt{}2h/g$$t+2(1/2gt2), simplifying to ΔY=g$$\sqrt{}2h/g$$t+gt2. I tried to set the two equations equal to each other, getting Vibt+1/2gt2=g$$\sqrt{}2h/g$$t+gt2. From here I'm not sure what to do - I tried to find what t was equal to, but Vib and the two degrees of t's (t and t2) are making it rather difficult. I've tried rearranging twice and got a different answer each time. Any suggestions on what I'm doing wrong, or maybe where to start for rearranging?

4. Oct 28, 2009