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 across the canyon is 63.6 m. If he desires a 3.0-second flight time...

(A) 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

Given,

h = 15 m ; R = 63.6 m ; t = 3 s

a)We know that the range is given by:

R = Vx t => Vx = R/t

Vx = 63.6/3 = 21.2 m/s

from eqn of motion

S = ut + 1/2 at^2

-15 = Vy 3 - 1/2 x 9.8 x 9

Vy = 9.7 m/s

theta = tan^-1(Vy/Vx) = tan^-1(9.7/21.2) = 24.6 deg

Hence, theta = 24.6 deg

b)V = sqrt (Vx^2 + Vy^2)

V = sqrt(21.2^2 + 9.7^2) = 23.31 m/s

Hence, V = 23.31 m/s

c)at langing,

Vy' = Vy + at

Vy' = 9.7 - 9.8 x 3 = -19.7 m

theta = tan^-1(-19.7/21.2) = -42.9 deg

Hence, theta = 42.9 deg (below horizontal)

d)V = sqrt (21.2^2 + -19.7^2) = 28.9 m/s

Hnece, V = 28.9 m/s

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