Posted by **Andy** on Friday, September 11, 2009 at 11:29pm.

A Hollywood Daredevil plans to jump a canyon 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 69.6 m across. If he desires a a 3.2 second flight time

a. what is the correct angle of his launch ramp?

b.what is his correct launch speed?

c.what is the correct angle for the landing ramp?

d.what is his predicted landing velocity?

- physics -
**drwls**, Saturday, September 12, 2009 at 1:19am
a. The 3.2 s flight time tells you what the vertical launch component, Vy, must be.

Vy*3.2 s - g*(3.2)^2/2 = -15

Vy = 7.87 m/s

The width of the canyon and the flight time tell you what the horizontal velocity component, Vx, must be

Vx*3.2s = 69.6 m

Vx = 21.75 m/s

The ratio Vy/Vx is the tangent of the launch angle.

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

c. Calculate the final vertical velocity at landing. It will be larger than the launch value of Vy because of the loss of height. Vy(final)/Vx will be the landing ramp angle tangent.

d. You can get this easily from the launch velocity and the 15 meter drop in altitude. V^2 increases by 2 g H, where H = 15 m.

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