The determined Wile E. Coyote is out once more to try to capture the elusive roadrunner. The coyote wears a new pair of Acme powered roller-skates, which provide a constant horizontal acceleration of 13 m/s2. The coyote starts off at rest 50 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 to reach the cliff before the coyote.

b) If the cliff is 140 m above the base of the canyon, find where the coyote lands in the canyon. Assume his skates are still in operation while he is in "flight" and that his horizontal component of acceleration remains constant.
(horizontally from the edge of the cliff)

To find the minimum speed the roadrunner must have to reach the cliff before the coyote, we can use the equations of motion.

a) First, let's calculate the time it takes for the coyote to reach the edge of the cliff.

We can use the equation:
distance = initial velocity * time + (1/2) * acceleration * time^2

Since the coyote starts from rest, the initial velocity is 0.
The distance to the cliff is 50 m.
The horizontal acceleration of the coyote is 13 m/s^2.

Plugging in these values, we get:
50 = 0 * t + (1/2) * 13 * t^2

Simplifying the equation, we get:
50 = (1/2) * 13 * t^2

Dividing both sides of the equation by (1/2) * 13, we get:
t^2 = 50 / ((1/2) * 13)

t^2 = 100 / 13

Taking the square root of both sides, we get:
t = sqrt(100 / 13)

Now that we know the time it takes for the coyote to reach the edge of the cliff, we can find the minimum speed the roadrunner must have to reach the cliff before the coyote.

The roadrunner needs to cover a distance of 50 m in this time.
So, the minimum speed required is given by:
speed = distance / time
speed = 50 / t

Plugging in the value of t, we get:
speed = 50 / sqrt(100 / 13)

Simplifying this expression gives us the answer to part a).

b) To find where the coyote lands in the canyon, we need to calculate the time it takes for the coyote to fall from the cliff to the canyon floor.

We can use the equation:
distance = initial velocity * time + (1/2) * acceleration * time^2

In this case, the coyote falls vertically, so the vertical acceleration is equal to the acceleration due to gravity, approximately 9.8 m/s^2.
The initial velocity in the vertical direction is 0 because the coyote starts falling from rest.
The distance from the cliff to the canyon floor is 140 m.

Plugging in these values, we get:
140 = 0 * t + (1/2) * 9.8 * t^2

Simplifying the equation, we get:
140 = (1/2) * 9.8 * t^2

Dividing both sides of the equation by (1/2) * 9.8, we get:
t^2 = 140 / ((1/2) * 9.8)

t^2 = 28.6

Taking the square root of both sides, we get:
t = sqrt(28.6)

Now that we know the time it takes for the coyote to fall, we can calculate the horizontal distance traveled by the coyote.

The horizontal distance is given by:
distance = initial velocity * time + (1/2) * horizontal acceleration * time^2

In this case, the initial velocity is 0 because the coyote starts falling from rest.
The horizontal acceleration is still 13 m/s^2.
The time we calculated earlier is used here.

Plugging in these values, we get:
distance = 0 * t + (1/2) * 13 * t^2

Simplifying the equation, we get:
distance = (1/2) * 13 * t^2

Plugging in the value of t, we get:
distance = (1/2) * 13 * (sqrt(28.6))^2

Simplifying this expression gives us the answer to part b).