Question

Wile E. Coyote is once again pursuing the Roadrunner, chasing the bird in a rocket-powered car. Unsurprisingly, the Roadrunner outsmarts him, and he sails off a cliff in the desert. His velocity when he leaves the cliff is horizontal with a magnitude of 24m/s, but the rocket continues to provide a constant horizontal acceleration of 3m/s2 . The cliff is 29 meters tall. How far from the base of the cliff does the coyote crash into the ground? Assume the ground is flat.

Answer 1

Answer:

The coyote crashes 66 m from the base of the cliff.

Explanation:

Hi there!

The equation of the position vector of the Coyote is the following:

r = (x0 + v0 · t + 1/2 · a · t², y0 + 1/2 · g · t²)

Where:

r = postion vector of the Coyote at time t.

x0 = initial horizontal position.

v0 = initial horizontal velocity.

t = time.

a = horizontal acceleration.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Let's place the origin of the frame of reference at the edge of the cliff so that x0 and y0 = 0.

When the Coyote reaches the ground, the vertical component of its position vector (r1 in the figure) will be equal to -29 m. When the vertical component of the position vector is -29 m, the horizontal component will be equal to the horizontal distance traveled by the Coyote (r1x in the figure). So, let's find the time at which the y-component of the position vector is -29 m:

y = y0 + 1/2 · g · t²  (y0 = 0)

-29 m = -1/2 · 9.8 m/s² · t²

t² = -29 m / -4.9 m/s²

t = 2.4 s

Now, let's find the x-component of the vector r1 in the figure:

x = x0 + v0 · t + 1/2 · a · t²     (x0 = 0)

x = 24 m/s · 2.4 s + 1/2 · 3 m/s² · (2.4 s)²  

x = 66 m

The coyote crashes 66 m from the base of the cliff.