Question

A Hollywood daredevil plans to jump the canyon shown in the figure on a motorcycle. There’s is a 15 m drop and the horizontal distance across the canyon is 69.2 m. If he desires a 2.9 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?

What is his predicted landing velocity?

Answer 1

Y = Uyt - 0.5gt^2

-15 = u sin a*2.8 - 0.5*9.8*2.9^2

u sin a = 9.36 ---------(1)

X = Ux*t

69.2 = u cos a * 2.9

u cos a = 23.86 -------(2)

dividing both equations we have

tan a = 0.392

a = 21.4^o

u = 23.86 / cos 21.4 = 25.63 m/s

Vx = Ux = 25.63 cos 21.4 = 23.86 m/s

Vy = Uy - gt = 25.63 cos 21.4 - 9.8*2.9 = -4.5 m/s

landing velocity = sqrt(23.86^2 + 4.5^2) = 24.28 m/s

angle = tan^-1(4.5 / 23.86 ) = 10.68^o below the horizontal

so the launch speed is 25.63 m/s

angle for launch ramp = 21.4^o

correct angle for landing ramp = 10.68^o below the horizontal

landing velocity = 24.28 m/s