- #1
FizzixIzFun
- 14
- 0
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.
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.
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.