# Homework Help: Reviewing physics I in my physics II class and having trouble

1. Sep 8, 2006

### FizzixIzFun

1)Wiley Coyote is chasing the roadrunner yet again. While running down the road, they come to a deep gorge, 15 m straight across and 78 m deep. The roadrunner launches itself across the gorge at a launch angle of 12 degrees above the horizontal, and lands with 2.4 m to spare. The acceleration of gravity is 9.81 m/s^2. What was the roadrunner's launch speed? Ignore air resistance. Answer in units of m/s.

2)Wiley Coyote launches himself across the gorge with the same initial speed, but at a lower launch angle. To his horror, he is short of the other lip by 0.8 m and falls into the gorge. What was Wiley Coyote's launch angle? Answer in units of degrees.

Ok, so I figured out #1. I got 20.48577875 m/s. My problem is on #2. I realized that his horizontal displacement would be 14.2 m and thus I had the equation 14.2=[Vcos(theta)][t] where V is the answer to #1. I solved that for t so I had t=14.2/[Vcos(theta)]. I then substituted that into (.5)(9.8)t^2 - [Vsin(theta)][t] - 78=0. I solved that for theta, but the value I found for theta was 83.9462306297 degrees. That can't be the right answer since Wiley launched himself at a "lower launch angle" than the roadrunner (whose launch angle was 12 degrees). I think that maybe this is one of those times where I would subtract the value of theta I found above from 90 degrees and that would give me my answer, but I'm just not sure. Your help would be extremely appreciated.

2. Sep 8, 2006

### Staff: Mentor

What's that "- 78" for? I assume that at time t the Coyote is at its original height (just 0.8 m short horizontally), not at the bottom of the gorge.

3. Sep 9, 2006

### FizzixIzFun

78 is the depth of the gorge so that you can figure out the amount of time it takes him to get from the point where he jumps to the point where he lands. it's from the kinematic x=x(initial) +V(initial)t + .5at^2

Last edited: Sep 9, 2006
4. Sep 9, 2006

### Staff: Mentor

Yes, but why are you using it in your equation? I assume that being "short of the other lip by 0.8 m" means that he returns to the starting level (y = 0, not y = -78) at x = 15 - 0.8.

(He doesn't land at the bottom of the gorge, he slams into the side!)

5. Sep 9, 2006

### FizzixIzFun

Ok, I'm lost. What should I do? Why doesn't he hit the bottom of the gorge?

Last edited: Sep 9, 2006
6. Sep 9, 2006

### Staff: Mentor

I thought I told you what to do? :tongue:

To understand what's going on, draw a picture. Like any other projectile, the coyote takes a parabolic path. To just miss making it to the other side means that he hits the wall of the gorge.

In order for him to hit the bottom (at x = 14.2), he would have to jump off at a much higher angle (as you figured out!). But that's not what happens. Instead, he just misses the edge, falling short by 0.8 m.

What if he overshot the distance by 0.8m instead of falling short? In that case he would have landed on the other side at a distance of x = 15.8 m from the start instead of at x = 14.2 m. (And his y coordinate when he landed would have been y = 0.) Make sense?

Big hint: Coordinates at start of motion: x = 0; y = 0. What are the coordinates at the point where he just passes the other edge? x = 14.2 m; y = 0!

(Note: I assume that "falling short" by 0.8 m means falling short horizontally by 0.8 m. But it could also mean falling short vertically by 0.8 m.)

7. Sep 9, 2006

### FizzixIzFun

Oh, I see. I need to use R=([V^2]sin(2[theta])) / g. So it would look like 14.2 = ([V^2]sin(2[theta])) / 9.81 where V is the answer to #1 and I would solve for theta. Right?

8. Sep 10, 2006

### Staff: Mentor

Sure, you could use the range formula to solve this. But your original method in your first post was just fine--except for the "-78".

Solve it both ways!