Wiley Coyote (Carnivouous hungribilous) is chasing the roadrunner (Speediburd cantcatchmi) yet again. While running down the road, they come to a deep gorge, 22 m straight across and 61 m deep. The...

Wiley Coyote (Carnivouous hungribilous) is chasing the roadrunner (Speediburd cantcatchmi) yet again. While running down the road, they come to a deep gorge, 22 m straight across and 61 m deep. The roadrunner launches itself across the gorge at a launch angle of 19◦ above the horizontal, and lands with 2.5 m to spare. The acceleration of gravity is 9.81 m/s 2 . What was the roadrunner’s launch speed? Ignore air resistance. Answer in units of m/s.

part two 

Wiley Coyote launches himself across the gorge with the same initial speed, but at a lower launch angle. To his horror, he is short of the other lip by 1.2 m and falls into the gorge. What was Wiley Coyote’s launch angle? Answer in units of ◦ .

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gsenviro | College Teacher | (Level 1) Educator Emeritus

Posted on

Let us consider the roadrunner and Wiley coyote as projectile, launched with an initial velocity and a certain angle with the horizontal. The range for a projectile is given as:

`R = (u^2 sin2theta)/g`

where, R is the range, u is the initial velocity, g is the acceleration due to gravity and `theta`  is the angle with respect to the horizontal. 

In case of roadrunner, the range = 22 m + 2.5 m = 24.5 m, `theta`  = 19 degrees and g = 9.81 m/s^2

Using the equation of range, we get:

`24.5 = (u^2 sin(2xx19))/9.81`

solving this equation, we get, u = 19.76 m/s.

In case of the coyote, range = 22 -1.2 m = 20.8 m

Using the same equation of range, with the same initial velocity, we get:

`20.8 = (19.76^2 sin2theta)/9.81`

solving this equation, we get, `theta`  = 15.75 degrees.

Hope this helps.

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