Pl e a s e h e l p y a l o c a l h u m a n o u t < 3 the famous looney toons character, wiley coyote, has missed the elusive road runner once again. this time, he speeds off the edge of a cliff at 10.0 m/s horizontal velocity. a) if the canyon is 300 m deep, how far from the base of the cliff does the coyote land? b) how fast would wiley coyote have to move horizontally off the edge of a cliff that was 100 meters high assuming he wanted to land the same distance away from the cliff as he did from a)

We are given that the speed Wiley the Coyote runs off the cliff. It is 10.0m/s. We are also given the height of the canyon, 300m. From this we can calculate the amount of time for Wiley to hit the bottom.

Consider only the y-component in this case. We can use the equation:

y = v*t + 0.5at^2

v in this case is NOT the velocity in the x direction, but the y direction. Since Wiley is going off the cliff at a horizontal velocity, his y velocity is 0. Thus v = 0. We can also substitute our acceleration due to gravity, g (9.81), in for a. Finally Wiley hits the floor which is 300 m. This means he traveled 300 m. We can plug that in for y. This leaves us with:

300 = 0.5 * 9.81 * t^2

Solving for t we get:

t = 7.82061887 s

Now, for the horizontal component to find how far he traveled. Assuming there is no air resistance, his velocity will be constant (Newton's First Law). Using the same equation as before, we can solve for distance. Since Wiley's not accelerating in the x direction,   a = 0. Thus leaving us with

x = v * t + 0.5 * 0 * t^2

x = v * t

Plug in the 10m/s and t from before we get:

x = (10m/s)*(7.82s) = 78.2061887 m

Now for Part B:

Given that the cliff is now 100 meters high, we can do the same thing we did last time. Solve for the time Wiley will take to hit the ground.

100 = 0.5 * 9.81 * t^2

t = 4.51523641

Now, we have the distance from part A ( 78.2061887). We can now solve for the velocity in the x direction:

78.2061887 = v * 4.515

v = 17.32050808 m/s

Hope this helps!

that would be because of both gravity and friction. so     c