# Projectile Motion of roller skates

1. Nov 1, 2006

### LastBloomingFlower

4) The determined Wile E. Coyote is out once more to try to capture the elusive roadrunner. The coyote wears a new pair of Acme power roller skates, which provide a constant horizontal acceleration of 15 m/s2, as shown in Figure P3.73. The coyote starts off at rest 70 m from the edge of a cliff at the instant the roadrunner zips by in the direction of the cliff.

(a) If the roadrunner moves with constant speed, find the minimum speed the roadrunner must have in order to reach the cliff before the coyote.

b) If the cliff is 100 m above the base of a canyon, find where the coyote lands in the canyon. (Assume that his skates are still in operation when he is in "flight" and that his horizontal component of acceleration remains constant at 15 m/s2.)

I don't understand even how to begin it...

2. Nov 1, 2006

### OlderDan

Begin by writing the equations of motion for each animal. Which one starts from rest? Which one has acceleration? What happens in the vertical direction to Mr. Coyote?

3. Nov 1, 2006

### geoffjb

Part A works in only one dimension (let's call it x). The roadrunner has constant non-zero velocity (zero acceleration), which implies an initial non-zero velocity. Coyote has an initial velocity of zero, but uniform acceleration. Find equations that will plot out exactly what it will take for both Coyote and Roadrunner to have the same position at the lip of the cliff.

Part B involves only Coyote. Break his velocity into y- and x- components. Use the former to determine how long it will take before he hits the ground, and then plug this time into an equation along with the latter to solve for his distance traveled.

4. Nov 1, 2006

### LastBloomingFlower

Well I have for the coyote
t=3.06s
A= 15 m/s/s
Vf=45.9
Vi=0
d=70

t=3.06
a=0
Vi=?
Vf= ?
d=70m

Am i doing it right so far?

5. Nov 1, 2006

### geoffjb

That all seems reasonable so far. Make it a habit to include units everywhere (e.g. on your final velocity, displacement, and time above).

6. Nov 1, 2006

### LastBloomingFlower

Geoffjb, have you worked out this problem at all? Just wondering. :)

7. Nov 1, 2006

### LastBloomingFlower

For the second part I found the time to be 1.71s and Vf to be 71.42 m/s and Vi to be 45.83m/s all of this beinbg in the y-component. So the x-component would have t=1.17s and A= 0 but how would I find d or any velocities?

8. Nov 2, 2006

### OlderDan

The problem says you should assume the horizontal acceleration remains constant for the whole time. You could do the problem either way, but I assume you want the answer to the problem as given. You know how long it took to reach the edge of the cliff, and you know the time in the air. The total time is the sum of the two. The same equation that worked for the x motion on top of the cliff keeps working the whole time.