A Hollywood daredevil plans to jump the canyon shown in the figure on a motorcyc

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 62.4 m across. If he desires a 3.0-second flight time..

1.what is the correct angle for his launch ramp (deg)?   2.50*10^1 deg   (correct)

2.What is his correct launch speed?  2.30*10^1 m/s   (correct)

3.What is the correct angle for his landing ramp (give a positive angle below the horizontal)?

                 I've already tried 20.8 deg

                                               36.16 deg

                                               -36.16 deg

                                               36.2 deg                      and they were all incorrect, I only have 1 try left

4.What is his predicted landing velocity. (Neglect air resistance.)

please answer only 3 and 4.

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 62.4 m across. If he desires a 3.0-second flight time.. what is the correct angle for his launch ramp (deg)? 2.50*10^1 deg (correct) What is his correct launch speed? 2.30*10^1 m/s (correct) 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 = 62.4m / 3.0s = 20.8 m/s
Also, s = -15 m = Vy*t +

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