Physics 12 Kinematics
posted by Oscar .
I'm not really sure how to use the algebra to prove the monkey shouldn't have dropped. Obviously if the hunter aimed the gun directlt at the monkey it wouldn't hit it as gravity would pull it down.
A huntsman fires a gun pointing at a monkey hanging on the branch of a tree. The huntsman and the base of the tree are on a flat horizontal. At the moment the huntsman fires the gun, the monkey will release it's grip on the branch of the tree. Use the initial bullet velocity Vo, horizontal distance Dx, monkey height Dy, and inital trajectory angle above.
a)Show the monkey shouldn't have dropped
b)For what minimum bullet velocity? Vo = (Dx/cos0)(sqrt(g/2Dy))
So I need to show that the monkey shouldn't have dropped when the bullet was fired somehow.... And then I have to find what minimum bullet velocity.
Thank you so much guys..

I have to do this with only variables...

a. the distance the monkey drops in time t, is the same distance the bullet drops in time t. You don't have to know t to know that.
b. you have the formula there.
That velocity is the velocity it takes for the bullet to reach impact directly below the monkey. 
But isn't the bullet going up ? I'm so confused on how to write this with variables... So Dy = T x9.8?

If there was no gravity, the monkey would not fall, nor the bullet fall from its path. But gravity exists, and both fall from the tree, and the straight line path, at the same rate.

I know that ss the bullet passes through the air, it will arc down due to gravity. If the monkey held the branch, and the hunter was far enough away, the bullet might pass beneath him, but because the monkey is falling while the bullet is travelling, he's going to get shot.... But how do I show this with algebra?

And thank you so much for taking the time to help me.

easy.
the impact will be directly below the monkey orig position, figure how long it takes the bullet to go that horizonal distance (time=ditance/horizonalcompoentvelocity)
Now, using that time, show the y position of the bullet is exactly the original height of the monkeyhow far it fell.
have fun, it's just algebra. 
Bullets horizontal displacement is Dx = (VoCos0)((VoSin0/9.8m/s^2x)2)
I'm not sure how to fund the y position of the byllet is exactly the original height of the monkey  how fat it fell..? 
The bullet must go horizontally a distance, Dx, so Dx = Vh / t or t = Vh / Dx = VcosA / D and
Vup = VsinA + gt .... or t = (Vup VsinA) /g .... sub
VcosA /D = (Vup VsinA) /g .... V = D(Vup VsinA)(g cosA)
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