I don't know how to even approach this... It's suppose to be on my physics test.
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..
To show that the monkey shouldn't have dropped when the bullet was fired, we need to consider the motion of both the monkey and the bullet.
Let's assume that the bullet and the monkey are both in freefall motion after being released.
The time it takes for the bullet to reach the height of the monkey is given by the equation:
t = sqrt(2Dy/g)
During this time, the horizontal distance the bullet travels is given by:
Dx = Vo * cosθ * t
Now, for the monkey to drop safely, we need to ensure that the bullet does not reach the height of the monkey before it is able to drop.
So, if the bullet reaches the height of the monkey (Dy) before the time it takes for the monkey to drop, the monkey will be hit.
Now, let's consider the vertical motion of the monkey. The time it takes for the monkey to fall from the branch is given by:
t_monkey = sqrt(2Dy/g)
If the bullet reaches the height of the monkey (Dy) before the time it takes for the monkey to fall, then the monkey will be hit.
So, to show that the monkey shouldn't have dropped, we need to compare the time it takes for the bullet to reach the height of the monkey (t) with the time it takes for the monkey to fall from the branch (t_monkey).
If t is greater than t_monkey, it means that the bullet will hit the monkey before it has a chance to drop. Therefore, the monkey shouldn't have dropped.
Now, to find the minimum bullet velocity required for the monkey to safely drop, we need to set the time it takes for the bullet to reach the height of the monkey (t) equal to the time it takes for the monkey to fall from the branch (t_monkey).
Substituting the equations for t and t_monkey, we get:
sqrt(2Dy/g) = sqrt(2Dy/g)
Next, we have the equation for the horizontal distance the bullet travels, Dx:
Dx = Vo * cosθ * sqrt(2Dy/g)
Rearranging this equation, we can solve for the minimum bullet velocity, Vo:
Vo = (Dx / cosθ) * sqrt(g/2Dy)
So, the minimum bullet velocity required for the monkey to safely drop is given by Vo = (Dx / cosθ) * sqrt(g/2Dy).
I hope this helps you understand how to approach this problem. Good luck with your physics test!