# Homework Help: Help With Projectile Motion

1. Oct 24, 2009

### AstraeaSophia

I think I'm on the right track for this problem, but just want to re-check my work.

1. The problem statement, all variables and given/known data
You are on the top of a 100m high (approximately 30 story) building. You just finished building a brick grill so the tenants of the building can have barbecues. You have one brick left and do not want have to lug it back off the roof. Luckily, you see a garbage truck with its roof open (top of trash pile is 2m above the road) coming down a road that passes by the building. The road is 30m from the building at closest approach. The truck is moving down the road with constant velocity 15m/s. You need to make sure you hit the trashpile, or someone is going to get hurt. You can throw the brick comfortably at an angle of 45 degrees above the horizontal.

a) What velocity do you need to throw the brick in order to make it to the top of the trashpile when the trash truck is 30m away?

2. Relevant equations
We're going over vector components of projectile motion, so we've got some new equations in the text that describe those components. When drawing the diagram for the angle of the initial throw, I'm using that to calculate the initial velocity required to make it 30m out. I used:

vy = sin$$\Theta$$(v0)
vx = cos$$\Theta$$(v0)
$$\Delta$$y = .5g($$\Delta$$x2/vx2)

We've also been introduced to dot products and cross products, which I must confess I don't entirely understand yet because they're so new. Not sure if they're required for this problem yet.

3. The attempt at a solution

I used the equation for the change in y-position and plugged in 98m for the value. Plugged in 30m for the change in x-position. My value for vx = 6.7082.

Then I used the cos$$\Theta$$ equation, plugged in my found value of vx, and eventually came out to the initial velocity needing to be 9.4868 m/s.

Is this the right approach? Just wanna double-check myself before I go any further. Thanks!

Last edited: Oct 24, 2009
2. Oct 24, 2009

### AstraeaSophia

Bumped because I added a different attempt to my problem description. Thanks!

3. Oct 24, 2009

### cepheid

Staff Emeritus
No part of your attempt takes into account the motion of the truck. Here is how I would approach this problem:

1. How long is it going to take the truck to arrive at a position directly underneath the thrower?

2. In order for the brick to travel upwards, reach a maximum height, begin falling, and reach the ground all in an amount of time equal to what was calculated in (1), what must its initial velocity be?

4. Oct 24, 2009

### AstraeaSophia

I'm not sure how I'd do it your way. You're asking me to first solve for how long it would take the truck to be right in front of the building so that it can catch the brick. What am I supposed to use as a reference point to find this measurement? Seems to me you have to find the time it takes the brick to fall first, then see how far back the truck has to be at the time of launch in order for it to be right in front of the building when the brick lands, which is actually what Part B of the problems asks (although I didn't list it). The prof seems to be asking the question in this order for a particular reason. First, I calculated the velocity for how hard the guy would have to throw the brick, then I used the equations

y(t) = y0+v0t+.5at2
$$\Delta$$x = v0$$\Delta$$t

and plugged in my initial velocity found in Part A. The first equation gave me the time it took the brick to reach the ground. Isn't this the correct approach? I used -98 m as my y-position and found a quadratic equation whose positive t = 5.54375 s.

I then plugged in the t I found into the $$\Delta$$x equation to find how far the truck would travel in that time period. The answer would be how far back on the road you'd have to be to end up directly underneath the brick at the time it reached "ground level." I found that position to be 83.1562 m if the truck is traveling at 15 m/s.

It seems like the speed of the truck isn't necessary to answer the first question, but is included so that the second part can be answered as a result of the first answer. Does this seem right to you guys?

Last edited: Oct 24, 2009
5. Oct 24, 2009

### cepheid

Staff Emeritus
I misinterpreted the 30 m as being the distance of the truck *down the road*, as opposed to, "away from the base of the building."

6. Oct 25, 2009

### AstraeaSophia

Oh, awesome. Thanks for the feedback!