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 68.2 m across. If he desires a 3.0-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

R = Vx*T

Vx = 68.2/3 = 22.733333 m/s


-Y = X*tan(theta) - 0.5*g*X^2/Vx^2

-15 = 68.2*tantheta - 4.9*68.2^2/22.733^2

theta = 23.1


a) V = Vx/cos(theta)= 24.71 m/s

b) theta = 23.1 degrees

c)

Vyf = Vy - gT = V*sin12.5 - gT = 20.337

angle = tan^-1(Vyf/Vx) = 41.81 degrees

Vf = sqrt(Vyf^2+Vx^2) = 30.50 m/s

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