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 66.6 m across. If he desires a 3.2-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

Part A)

In the y direction, the formula to apply = d = vot + .5at^2

d = -15m, so..

-15 = vo(3.2) + (.5)(-9.8)(3.2^2)

vo = 10.99 m/s in the y

In the x, apply d = vt

66.6 = v(3.2)

v = 20.81 m/s

Combine the two using the pythagorean theorem

v(net) = sqrt[(20.81)^2 + (10.99)^2]

v = 23.5 m/s

The angle of launch is found using the tan function (although this is not asked for, here it is anyway)

tan(angle) = 10.99/20.81

angle = 27.84 degrees  

Part B)

The angle of landing...

vf in the x does not change

vf in the y is found using

vf = vo + at

vf = (10.99) + (-9.8)(3.2)

vf = -20.37 m/s (negative since it is downward)

The angle is found from the tangent function

tan angle = -20.37/20.81

angle = -44.4 degrees

That is 44.4 degrees below the horizontal

Part C)
Net landing velocity = sqrt [(20.37)^2 + (20.81)^2]

Velocity = 29.1 m/s at a 44.4 degree angle below the horizontal

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