A Hollywood daredevil plans to jump the canyon shown in the figure on a motorcyc
ID: 2272562 • Letter: A
Question
A Hollywood daredevil plans to jump the canyon shown in the figure on a motorcycle. There is a 15. m drop and the horizontal distance originally planned was 60. m but it turns out the canyon is really 61.1 m across. If he desires a 3.2-second flight time, what is the correct angle for his launch ramp (deg)?
What is his correct launch speed?
What is the correct angle for his landing ramp (give a positive angle below the horizontal)?
What is his predicted landing velocity. (Neglect air resistance.)
A Hollywood daredevil plans to jump the canyon shown in the figure on a motorcycle. There is a 15. m drop and the horizontal distance originally planned was 60. m but it turns out the canyon is really 61.1 m across. If he desires a 3.2-second flight time, what is the correct angle for his launch ramp (deg)? What is his correct launch speed? What is the correct angle for his landing ramp (give a positive angle below the horizontal)? What is his predicted landing velocity. (Neglect air resistance.)Explanation / Answer
The horizontal velocity should be Vx = 62.3m / 3.1s = 20.1 m/s
Also, s = -15 m = Vy*t +
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.