A monkey hangs from a branch at a height, h, above the hunter A hunter stands on
ID: 1627982 • Letter: A
Question
A monkey hangs from a branch at a height, h, above the hunter A hunter stands on the ground at a distance, d, from the tree, and is pointing his gun at the monkey at an angle, , to the horizontal, as shown below. The hunter fires his gun at an initial velocity of, Vo. The monkey releases his grip and starts to fall at exactly the same time as the bullet leaves the nuzzle of the gun.
1. Write the equation of motion for the bullet and the equation of motion of the monkey in terms of Vo, d, h, t, , and g. You should have four equations, two for the monkey's motion and twofor the bullet's motion
2. What must the velocity of the bullet be greater or equal to in order for the bullet to always hit the monkey? (Your answer wil be an equality in terms of Vo, d, h, g and .)
3. Use your result from question 2 to determine the minumum velocity of the bullet to hit the monkey if h = 10m and d = 20m.
4. Under what condition will the bullet hit the monkey for any initial velocity of the bullet greater than or equal to zero?
.c----- 1;sExplanation / Answer
First we have to set up the scenario.
You have a man aiming at the monkey which makes a triangle with a width of d, and a height of h. The hypotenuse of this triangle is sqrt(d^2 + h^2).
The x component of the velocity is vo*cos or vod/root(d^2 + h^2)
The bullet's position equation in the x direction is:
X(t) = 1/2 a t^2 + Vx t + Xo (Xo and a are zero because you start at the origin and there is no acceleration)
X(t) = Vx t
d = Vx t
Substituting the value we came up with for Vx earlier and solving for time gives:
d = vot (d/root(d^2 + h^2))
(d/vo)(root(d^2 + h^2)/d) = t
t = root(d^2 + h^2)/vo
Now, we consider the monkey and it's position equation:
Y(t) = 1/2 g t^2 + Vy t + h (Vy is zero)
Y(t) = 1/2 g t^2 + h
Substitute the value we found earlier for how long it takes the bullet to travel d:
Y(t) = 1/2 g [root(d^2 +h^2)/vo]^2 + h
Y(t) = 1/2 g (d^2 + h^2)/vo^2 + h
The delta-Y will be the difference between Y at t = 0 and Y at impact:
delta-Y = Y(0) - Y(t)
delta-Y = h - (1/2 g (d^2 + h^2)/vo^2 + h) (h's cancel out)
delta-Y = 1/2 g (d^2 + h^2)/vo^2