Wile E. Coyote (Carnivorus hungribilus) is chasing the Roadrunner (Speedibus can

Question

Wile E. Coyote (Carnivorus hungribilus) is chasing the Roadrunner (Speedibus cantcatchmi) yet again. While running down the road, they come to a deep gorge, 14.4 m straight across and 100 m deep. The Roadrunner launches himself across the gorge at a launch angle of 13o above the horizontal, and lands with 1.3 m to spare. (a) What was the Roadrunner's launch speed? x m/s (b) Wile E. Coyote launches himself across the gorge with the same initial speed, but at a different launch angle. To his horror, he is short of the other lip by 0.50 m. What was his launch angle? (Assume that it was less than 13°.) Sketch the situation. In the absence of air resistance, the accelerations of both Wile E. Coyote and the Roadrunner are constant and you can use constant-acceleration equations to express their coordinates at any time during their leaps across the gorge. By eliminating the parameter t between these equations, you can obtain an expression that relates their y coordinates to their x coordinates which you can solve for their launch angles

Explanation / Answer

Given,

theta = 13 deg ; x = 1.3 m ; L = 14.4 m ; d = 100 m

a)We know that the horizontal distance covered by the projectile(range)is given by:

R = v0^2 sin(2theta)/g

solving for v0 we get

v0 = sqrt [Rg/sin(2theta)]

R = 14.4 + 1.3 = 15.7 m

2theta = 2 x 13 = 26 deg

v0 = sqrt (15.7 x 9.8/sin26) = 18.73 m/s

Hence, launch speed = v0 = 18.73 m/s

b)For Coyote the Range is:

R = 14.4 - 0.5 = 13.9 m

from the eqn of range, solving for theta we get

theta = 1/2 sin^-1 (Rg/v0^2)

theta = 0.5 x sin^-1(13.9 x 9.8/18.73^2) = 11.42 deg

Hence, theta = 11.42 deg

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