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Sep6-11, 08:48 PM   #1
 

Projectile Motion: Finding the correct angle for a launch


1. The problem statement, all variables and given/known data
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 69.4 m across. If he desires a 3.0-second flight time, what is the correct angle for his launch ramp (deg)?


2. Relevant equations
vx = vxo

x − xo = (vox)t

(vy) = (vyo) − gt

y − yo = (voy)t −(1/2)g t^2

Range =((vo^2)/g) sin(2θ)

3. The attempt at a solution
x − xo = (vox)t
69.4m = (vox)(3s)
23.13 m/s = vox

Range =((vo^2)/g) sin(2θ)
69.4m ={ (23.13^2 m/s)/9.81 m/s^2 } sin(2θ)
θ = error
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