Mechanical Energy and Trajectory

[James98765]

Homework Statement

I am having difficulty with the concept of a certain question from my textbook that goes as follows:

Your friend's Frisbee has become stuck 16 meters above the ground in a tree. You want to dislodge the Frisbee by throwing a rock at it. The Frisbee is stuck pretty tight, so you figure the rock needs to be traveling at least 5.0 m/s when it hits the Frisbee. Does the speed at which you throw the rock depend on the angle at which you throw it? Explain.

(There is another mathematical part of this question but I have decided to disregard it to get straight to my question)

Homework Equations

The answer in the back of my textbook states 'No' which goes against my understanding.

The Attempt at a Solution

The problem I am having is I cannot possibly see how the launch velocity wouldn't depend on the launch angle. I understand that inorder for the rock to travel 16 meters vertically and collide with the Frisbee in the tree at 5.0 m/s, the final kinetic energy of the rock must be at a certain value no matter the launch angle. But if the person throwing the rock is changing the launch angle then the initial velocity must change in ensure that the rock hits the Frisbee. If anyone could explain this question to me it would be much appreciated. Thanks you!

-James

 

[rock.freak667]

initial speed=v

initial horizontal velocity v_x=vcos$\theta$
initial vertical velocity v_y=vsin$\theta$

$$v_R=\sqrt{v_y^2+v_x^2}$$(v_R=5m/s at 16m)

v_x is constant throughout the motion.

Using v²=u²-2gs we have

$$v_y^2=v^2cos^2\theta-2g(16)$$

Now can you see why v_R does not depend on $\theta$?

 

[James98765]

Yes thank you. I think I am beginning to realize that because the rock must hit the tree at a specific velocity it therefore must have a specific final kinetic energy regardless of launch angle. Because the rock is traveling through a height of 16m it will always have the same change in kinetic energy therefore it must have a certain initial kinetic energy to provide the proper final kinetic energy. Thank you for your help!

-James

 

[ice2morrow]

I was actually wondering if it would be possible to get the other answer to this question. If you release the rock 1.80 above the ground, with what minimum speed (in m/s) must you throw it? I have no idea how to do this problem.

 

[James98765]

Sorry about the late response. I didn't see your post until now. I will explain the problem assuming you have an understanding of kinetic and potential energy.

Because we know the velocity at which the rock must hit the tree, we know the kinetic energy that it must have when it reaches the tree. That kinetic energy is as follows:

K_f = (1/2)MV_f² = (1/2)M(5.0 m/s)²

Assuming we don't know the mass of the rock we will let it remain variable M. Becuase the rock will be at it's maximum height during the time it hits the tree, we can assume it's final potential energy is:

U_f = Mgh_f = Mg(16 m)

It's intital potential energy is as follows:

U_i = Mgh_i = Mg(1.8 m)

Now the hard part is overwith. We know the change in potential energy (U_f - U_i) and we know the final kinetic energy so we know how much kinetic energy we must give the rock to allow it to reach the tree at 5.0 m/s:

K_i = dU + K_f
(1/2)MV_i² = Mg(h_f - h_i) + (1/2)MV_f²

Noticing that mass above cancels and solving for velocity:

V_i = [2g(h_f - h_i) + V_f]^(1/2)
V_i = 16.7 m/s

 

[ice2morrow]

Thanks a bunch!