What is the magnitude of launch velocity and angle of launch in the following case:Wiley Coyote is chasing the roadrunner yet again.  While running down the road, they come to a deep gorge, 25m...

What is the magnitude of launch velocity and angle of launch in the following case:

Wiley Coyote is chasing the roadrunner yet again.  While running down the road, they come to a deep gorge, 25m straight across and 121m deep.

The roadrunner launches itself across the gorge at a launch angle of 17 degrees above the horizontal, and lands with 2.2m to spare.  The acceleration of gravity is 9.81m/s^2.  What was the roadrunner's launch speed? Ignore air resistance. Answer in units of m/s

Wiley Coyote with the same initial speed, but at a lower launch angle.  To his horror, he is short of the other lip by 0.6m and falls into the gorge. What was Wiley Coyote's launch angle? answer in units of degrees.

Asked on by pavilion77

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The roadrunner is being chased by Wiley Coyote. When they reach the gorge, the road runner launches itself across it at an angle of 17 degrees with the horizontal and lands with 2.2m to spare.

Let the magnitude of the launch velocity be V. The horizontal component of the the launch velocity is V*cos 17 and the vertical component is V*sin 17.

The vertical velocity changes due to the acceleration due to gravity acting downwards and after a time T is equal to -V*sin 17. This is the time the roadrunner is in motion.

-V*sin 17 = V*sin 17 - 9.81*T

=> T = 2*V*sin 17/9.81

In this time it travels a horizontal distance of 25 + 2.2 = 27.2m

=> (2*V*sin 17/9.81)*V*cos 17 = 27.2

=> V^2*sin 34 = 27.2*9.81

=> V = 21.8 m/s

The Coyote launches at the same magnitude but the angle is different due to which the horizontal distance traveled is only 24.4 m

=> 21.8^2*sin x = 24.4*9.81

=> sin x = 0.7096

=> x = 45.21

The angle of launch is 22.60 degrees.

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