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 61.8 m across. If he desires a 3.2-second flight time, what is the correct angle for his launch ramp (deg)?

b. What is his correct launch speed?

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

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

Explanation / Answer

y = -15 m, x = 61.8 m, t = 3.2 s, launch speed = v
a) find the correct angle for his launch ramp

x = vcos*t, so vcos = x/t = 19.31

y = vsin*t - gt2/2, so vsin = y/t + gt/2 = 10.99

tan = 10.99/19.31 = 29.7o
b. his correct launch speed = v

v = 19.31/cos = 22.2 m/s

c. the correct angle for his landing ramp =

landing velocity = u

ux = vcos = 19.31

uy = vsin - gt = -20.4

= tan-1(20.4/19.31) = 46.6o

d. his predicted landing velocity = u

u = 28.1 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.