A)A Hollywood daredevil plans to jump the canyon shown in the figure on a motorc

ID: 2132832 • Letter: A

Question

A)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 68.4

m across. If he desires a 2.8-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.)

Part A)
In the x direction, d = vt

68.4 = v(cos angle)(2.8)

v(cos angle) = 60.14 m/s

In the y direction

d = vot + .5at^2

-15 = v(sin angle)(2.8) + (.5)(-9.8)(2.8^2)

v(sin angle) = 8.36 m/s

vsin(angle)/v(cos angle) = 8.36/60.14

tan(angle) = 8.36/60.14

angle = 7.92 degrees

Part B)

v(cos 7.92) = 60.14

v = 60.72 m/s

Part C)
Final vx = 60.14 m/s

Final vy is found from vf^2 = vo^2 + 2ad

vf^2 = 8.36^2 + 2(-9.8)(-15)

vf = 19.1 m/s

The angle is from the tangent function

tan(angle) = 19.1/60.14

angle = 17.6 degrees

Part D)
From the pythagorean theorem

v = sqrt[(19.1)^2 + (60.14)^2]

v = 63.1 m/s

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