# Some projectile motion problems

1. Apr 11, 2004

### sidm

Just two questions:

agent tim flying a constant 185km/h horizontally in a low flying helicopter wants to drop a small explosive onto a master cimnial's automobile travelling 145km/h on a level highway 88meters below. at what angle should the car be in his sights when the bomb is released.

The answer is listed as 61.8degrees. I cannot get this answer for the life of me....an answer and brief explanation would be very much appreciated.

another problem i'm having trouble with:

Romeo is chucking pebbles gently up to juliets window and he wants hte pebbles to hit the window with only a horizontal component of velocity. he is standing at the edge of a rose garder 8.0m below her window and 9.m from the base of the wall. how fast are the pebbles going when they hit her window?

Welll..again I can't get the answer..i could, however, if I knew the original angle it was thrown at...can I do this by using basic trig w/ the length-made triangle?

Thanks for the help.

2. Apr 11, 2004

### Chen

Show us what you've done so far, so we can correct your answers (rather than solve the problems for you).

3. Apr 11, 2004

### sidm

ok for the first problem:
y-component data:

the final y-velocity of bomb = squareroot(2(9.8)(88)) assuming initial velocity is zero.

x-velocity is constant (195000/3600)m/s

since y-velocity is opposite the desired angle and x-velocity is adjacent.
tanx = -41.53/51.38
x = 51.05 degrees.

that's the first. i'm lost on the 2nd.

4. Apr 11, 2004

### Chen

What you did was find the angle between the velocity of the bomb upon impact and the ground. That is not what the question asks for...

First of all you need to find the time it takes the bomb to hit the ground (I get 4.24s). The relative speed of the airplane and the car is 40km/h, so to make it easier let's imagine that the car is stationary, and the airplane is approaching it at 40km/h. So once the bomb is thrown, what horizontal distance will it cover before hitting the ground? And what is the relationship between that distance and the height of the airplane above the road?

5. Apr 11, 2004

### sidm

hahah..cool. Thanks a bunch Chen! Just started this stuff yesterday (self-teaching)..using the book: PHYSICS: GIANCOLI (5th ed) - some tough problems. Any chance you could help me w/ the 2nd problem. Ok..ill just ask. the velocities of each component are defined as vcosx or vsinx right? Would it be safe to assume that x is the angle that occurs in the right triangle w/ values 8 and 9? (48.37 degrees is what i get).

Thanks again.

6. Apr 11, 2004

### Chen

For the second problem, divide the motion into two axes: X and Y.

$$x(t) = {v_0}_xt$$
$$v_x(t) = constant$$
$$y(t) = {v_0}_yt - \frac{1}{2}gt^2$$
$$v_y(t) = {v_0}_y - gt$$

Do you understand why?

Now, you know that the vertical component of the velocity should be zero when the pebbles hit the window. Therefore:

$$v_y(t_f) = {v_0}_y - gt_f = 0$$
$$t_f = \frac{{v_0}_y}{g}$$

So now we have the total time it takes the pebble to reach the window. Let's plug that into x(t) and y(t), because we also know their values at the end of the throw:

$$x(t_f) = 9_m = {v_0}_x\frac{{v_0}_y}{g}$$
$$y(t_f) = 8_m = {v_0}_y\frac{{v_0}_y}{g} - \frac{1}{2}g\frac{{v_0}_y^2}{g^2} = \frac{{v_0}_y^2}{g} - \frac{1}{2}\frac{{v_0}_y^2}{g} = \frac{1}{2}\frac{{v_0}_y^2}{g}$$

And now you have two equations with two variables. Solve for v0x (that is what the problem asks for) and v0y. I get a horizontal velocity of 7.04 m/s.

Last edited: Apr 11, 2004
7. Apr 11, 2004

### Chen

Yes, that's how you define each component of the velocity. But x is not the angle between the distances as well. I will explain why but it doesn't have anything to do with the problem at hand.

The direction of speed at any moment is the derivative of the displacement, as you must have learned. Now, you know how the rocks travel in this problem right? They go in a parabola. And what happens to the direction of the velocity at each moment? It "decreases", or in other words the angle between the velocity and the horizon decreases, due to gravity. So if the initial direction of the velocity is parallel to the direction of the window, it would never get to it! Because right after it is thrown, the angle will decrease more and more. You would only throw the rock directly at the window if there was no acceleration, i.e gravity, because then the direction of the velocity of the rock would not change at all.

Can you understand my poor explanation?