# Wile E. Coyote is holding a “HEAVY DUTY ACME ANVIL” on a cliff that is 40.0 meters high. The Roadrunner (beep-beep), who is 1.0 meter tall, is running on a road toward the cliff at a constant velocity of 10.0 m/s. Wile E. Coyote wants to drop the anvil on the Roadrunner’s head. How far away should the Roadrunner be when Wile E. drops the anvil?

I got a time of 2.89, using the position in the y direction .. then multiplied by the 10 m/s velocity (x=vt) to get 28.9.. but the choices are 28.2 and 28.6??

1. The anvil falls 39m
39=1/2 g t^2
t=sqrt(78/9.8)

2. y = 39m
g = 9.8 m/s/s
vo = 0 m/s (dropped from rest)
y = 0.5gt^2
t = sqrt(39/4.9) = approx. 2.82s (rounded from 2.821202523)
So now you have t.
x = vt, and v = 10 m/s
x = 10(2.8) gets you 28.2 meters.

My homework says the answer should be 28.6, though.

3. y = 39m - 1.0 m (Remember the Roadrunner's height!)= 38m
g = 9.8 m/s/s
vo = 0 m/s (dropped from rest)
y = 0.5gt^2
t = sqrt(38/4.9) = approx. 2.78s (rounded from 2.784798384)
So now you have t.
x = vt, and v = 10 m/s
x = 10(2.78) gets you 27.8 meters.

My homework says the answer should be 28.6, though.

4. Actually, this is how I got it: (based on my worksheet)
dy=40.0m
= (40.0m/4.9)
= squareroot(8.16) = 2.857 round off = 2.86 (10m)
= 28.6 m