1)Wiley Coyote is chasing the roadrunner yet again. While running down the road, they come to a deep gorge, 15 m straight across and 78 m deep. The roadrunner launches itself across the gorge at a launch angle of 12 degrees above the horizontal, and lands with 2.4 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.(adsbygoogle = window.adsbygoogle || []).push({});

2)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 0.8 m and falls into the gorge. What was Wiley Coyote's launch angle? Answer in units of degrees.

Ok, so I figured out #1. I got 20.48577875 m/s. My problem is on #2. I realized that his horizontal displacement would be 14.2 m and thus I had the equation 14.2=[Vcos(theta)][t] where V is the answer to #1. I solved that for t so I had t=14.2/[Vcos(theta)]. I then substituted that into (.5)(9.8)t^2 - [Vsin(theta)][t] - 78=0. I solved that for theta, but the value I found for theta was 83.9462306297 degrees. That can't be the right answer since Wiley launched himself at a "lower launch angle" than the roadrunner (whose launch angle was 12 degrees). I think that maybe this is one of those times where I would subtract the value of theta I found above from 90 degrees and that would give me my answer, but I'm just not sure. Your help would be extremely appreciated.

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# Homework Help: Reviewing physics I in my physics II class and having trouble

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