1) A Hollywood daredevil plans to jump the canyon shown in the figure on a motor
ID: 2022049 • Letter: 1
Question
1) 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 70.0 m across. If he desires a 3.2-second flight time, what is the correct angle for his launch ramp (deg)?2) What is his correct launch speed?
3) What is the correct angle for his landing ramp (give a positive angle below the horizontal)?
4) What is his predicted landing velocity. (Neglect air resistance.)
(please show work! thanks!)
Explanation / Answer
I just asked this question... so here's my question and my numbers a LITTLE different, so just substitute yours for mine 1. 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 69.7 m across. If he desires a 2.8-second flight time, what is the correct angle for his launch ramp (deg)? known y0 = 0 m y = -15m x = 69.7 m x0 = 0m t = 2.8s 1. find ? x = x0 +v0xt 69.7 m = 0 m + v0x(2.8 s) v0x = 24.9 m/s y = y0 + v0y t - 0.5 gt2 -15 m = 0m + v0y (2.8s) - 0.5 (9.8m/s2)(2.8s)2 v0y = 8.36 m/s so ? = tan-1 (v0y/v0x) ? = 18.6° 2 v0 = sqrt(v0x2 + v0y2) v0 = sqrt((24.9m/s)2 + (8.36m/s)2) v0 = 26.3 m/s 3 vx = v0x vx = 24.9 m/s vy = v0y - at vy = 8.36 m/s - (9.8m/s2) (2.8 s) vy = -19.0 m/s ? = tan-1(vy/vx) ? = tan-1(-19.0m/s / 24.9 m/s) ? = -37.3° so ? = 37.3° below the horizontal 4 v = sqrt(vx2 + vy2) v = sqrt( (24.9m/s)2 + (-19.0 m/s)2) v = 31.3 m/s
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