Question

Wiley Coyote has missed the elusive roadrunner once again. This time, he leaves the edge of the cliff at 50.0 m/s horizontal velocity. If the canyon is 100 m deep, how far from the edge of the cliff does the coyote land?

Ans: 226 m

(but please specify how that is)

Answer 1

Position:
y_inital = 100 m
x_inital = 0 m (we are choosing the starting value to bezero)

Velocity:
v_{x initial} = 50 m/s
v_{y initial} = 0 m/s

Acceleration:
a_x = 0 m/s² (we are neglecting airresistance)
a_y = g = -9.8m/s²

A Free Body Diagram couldn't hurt either:




I drew the velocity and acceleration components separate from thecoyote so it didn't get cluttered. The arc represents thepath of the coyote.

There are two parts to this question. First, we have todetermine how long it takes for the coyote to hit the bottom. Then, we can use that time to determine the distance the coyote hastraveled in the horizontal component.

One of the key points to understanding this problem isunderstanding that the horizontal and vertical components areindependent. That is, the acceleration due to gravity is onlygoing to change the velocity of the coyote in the verticaldirection. It will not affect the velocity of the coyote inthe horizontal direction.

So, the first part: Determine the amount of time that ittakes the coyote to hit the ground.

We can use the kinematics equations for this:

x = v_avgt   where v_avg = averagevelocity = (v_final-v_initial)/2
x = v₀t + 1/2 at²
v = v₀ + at
v² = v₀²+2ax

X is the displacement (change in position), v is the final velocity(except in the first formula), v₀ is the initialvelocity, a is acceleration, and t is time.

So, which one do we use (most easily)? Think about what weare looking for: time. That tells us to look at the topthree formulas because the fourth doesn't have a "t" in it. Now look at all of the other variables in the formulas and considerwhat we know and what we don't know. We know the distancethat the coyote will fall, the acceleration, and the initialvelocity in the vertical direction. We don't know the finalvelocity that the coyote will have (just before hitting theground). The second equation is the only one that isindependent of final velocity, so we'll use that.

x = v₀t + 1/2 at²

Changing the variables so as not to confuse it with the horizontalcomponent:

y = v_0yt + 1/2 a_yt²

Now we can substitute in what we know:

-100 m = (0 m/s)(t) + 1/2 (-9.8m/s²)(t²)

And solve for t:

-100 m = 1/2 (-9.8m/s²)(t²)

-200 m = (-9.8m/s²)(t²)

-200 m / (-9.8m/s²) = t²

√(-200 m / (-9.8m/s²) ) = t

√(-200 m / (-9.8m/s²) ) = t

4.52s = t

Now we know how long the coyote will be traveling in the horizontaldirection.

Using this time as well as the velocity in the horizontaldirection, we can determine how far (horizontally) the coyote willland from his starting point:

x = vt

x = (50 m/s)(4.52 s) = 226 m