Monkey and hunter - forces, vectors

[mgsk]

Monkey and hunter -- forces, vectors

Homework Statement

I'm currently supplementing my institutional education with some me-time study. I'm studying maths and physics A-levels (or, rather, A-level equivalent).

Yale University has an awesome "online courseware" which makes available to the public videos of lectures and study materials (problem sets, etc.). My problem refers to the problem set given for the Newton's Laws lectures, found here: problem set, solutions, specifically problem 8. The problem is quoted here:

A monkey is hanging from a height h and a person d meters away from the tree and on the ground, wants to zap it (in today's version with a tranquilizer gun and in the original version, a hunting rifle). He aims straight at the monkey and fires. This would of course work in the absence of gravity but show that it will work even in its presence provided the initial speed obeys \sqrt{\left ( d^2+h^2 \right ) \frac{g}{2h}}. What does this requirement ensure? Given that this will also work if a pulse of laser light is used, what do you learn about light in a gravitational field?

Homework Equations

I began with asking what are the positions for each object's motion. The monkey is easy. It's "x" position is unchanging, and so does not vary with time: x_m =d; while it's "y" position is changing with time: y_m = h-\frac{1}{2} g t^2. Next, and this is where my approach differs to that given in the solution, I define the components of the bullet. The bullet begins with some initial velocity v_0 in the horizontal direction, and so by my understanding of forces and Newtonian mechanics, this horizontal velocity is constant (ignoring air resistance), so that the bullet's "x" position is: x_b=v_0 t; and like the monkey, the bullet's "y" position changes because of gravity: y_b=h-\frac{1}{2} g t^2.

The Attempt at a Solution

My reasoning is then: for the bullet to hit the monkey, the bullet must travel a horizontal distance d in a time that is no greater than the time it takes the monkey to fall a distance h; and because I'm dealing with the horizontal distance, I can focus on the "horizontal" component of the bullet's trajectory. Thus, the time for the bullet to travel a distance d is: t_{b}=\frac{d}{v_0}; and the time it takes for the monkey to reach the ground is (given by solving y_m=0 for t): t_m=\sqrt{\frac{2h}{g}}. Then equating the two and solving for v_0: v_0=\frac{d\sqrt{g}}{\sqrt{2h}}.

Can someone spot the error in my reasoning? Many thanks!

P.S. Sorry for the poor formatting. I'm not familiar with this forum's LaTeX tools.

 

[haruspex]

My problem refers to the problem set given for the Newton's Laws lectures, found here: problem set, solutions, specifically problem 8. The problem is quoted here:

They seem to have forgotten to mention that on hearing the shot the monkey let's go of the branch (and the bullet is much slower than the speed of sound). (And you lost the "v >" in quoting the condition.)

My reasoning is then: for the bullet to hit the monkey, the bullet must travel a horizontal distance d in a time that is no greater than the time it takes the monkey to fall a distance h; and because I'm dealing with the horizontal distance, I can focus on the "horizontal" component of the bullet's trajectory.
v_0=\frac{d\sqrt{g}}{\sqrt{2h}}

... which is the correct answer for what you focused on - the horizontal component. But the question posed wants the total speed.

 

[Poetria]

I have a similar question: If somebody wants to shoot an object falling from a cliff with a parachute attached, where should he aim?
There is a drag force involved and that's why I think it is better to aim above the package. Drag force increases proportionally to velocity squared.
Am I right?

 

[haruspex]

I have a similar question: If somebody wants to shoot an object falling from a cliff with a parachute attached, where should he aim?
There is a drag force involved and that's why I think it is better to aim above the package. Drag force increases proportionally to velocity squared.
Am I right?

To figure out where to aim you would need to solve the drag equation. If it is directly proportional to the square of the speed then you can certainly find the velocity as a function of time (need to integrate a secant), but as I recall, the second integral to find the displacement by time can only be done numerically.
In reality, it's more like a general quadratic, involving a linear term.

The bullet also suffers drag. This is way more complicated because it is not moving vertically.

 

[J Hann]

What the question may be asking is to show that
h - 1/2 g t^2 > 0
Doesn't this have to be true if the bullet is to reach the monkey before the monkey reaches the ground?
Now use this to eliminate t from the equation using the given conditions.

 

[haruspex]

What the question may be asking is to show that
h - 1/2 g t^2 > 0
Doesn't this have to be true if the bullet is to reach the monkey before the monkey reaches the ground?
Now use this to eliminate t from the equation using the given conditions.

The original question was dealt with 4 years ago. The resurrection of the thread concerns drag. Probably should have been a new thread.

 

[J Hann]

In the monkey and hunter problem, the traditional explanation is that the bullet falls away
from the initial line of sight at the same rate that the monkey falls.
Thus, the bullet will hit the monkey.
Now, qualitatively, it should be clear that if the monkey falls at a slower rate than the
bullet then the bullet would pass under the monkey if the hunter was initially
aiming directly at the monkey.

 

[haruspex]

if the monkey falls at a slower rate than the
bullet then the bullet would pass under the monkey if the hunter was initially
aiming directly at the monkey.

Yes, that's what Poetria deduced in post #3.

 

[Poetria]

To figure out where to aim you would need to solve the drag equation. If it is directly proportional to the square of the speed then you can certainly find the velocity as a function of time (need to integrate a secant), but as I recall, the second integral to find the displacement by time can only be done numerically.
In reality, it's more like a general quadratic, involving a linear term.

The bullet also suffers drag. This is way more complicated because it is not moving vertically.

This question is a kind of thought experiment - too little data. At first, I thought it would be better to aim below the object with a parachute - I didn't take into account gravity acting on the arrow and I thought drag force would be zero at the instant of shooting as initial velocity is zero.

Sorry for having caused 'drag' on the forum. :) I was searching for similar situations to understand this better. :)