Wiley Coyote has missed the elusive road runner once again.This time,he leaves the edge of the cliff at 52.4 m/s horizontal velocity.If the canyon is 140 m deep,how far from his starting point at the edge of the cliff does the coyote land?
A)280 m
B)198 m
C)396 m
D)140 m
E)70.0 m

Question

Quizplus · Accepted Answer

The correct answer is A because the coyote's horizontal velocity is constant, so the horizontal distance he travels is equal to his velocity multiplied by the time he spends in the air. The time he spends in the air can be found using the equation of motion:$$y = vi*t + 1/2*a*t^2$$where y is the vertical displacement, vi is the initial vertical velocity, a is the acceleration due to gravity, and t is the time.In this case, the vertical displacement is -140 m, the initial vertical velocity is 0 m/s, and the acceleration due to gravity is -9.8 m/s^2. Solving for t, we get:$$t = 5.7 s$$Therefore, the horizontal distance the coyote travels is:$$d = v*t = 52.4 m/s * 5.7 s = 299.9 m$$Since the coyote starts at the edge of the cliff, the total distance he travels is 299.9 m + 140 m = 439.9 m.

Wiley Coyote Has Missed the Elusive Road Runner Once Again