Posted by Susan on .
A daredevil decides to jump a canyon of width 10 m. To do so, he drives a motorcycle up an incline sloped at an angle of 15 degrees. What minimum speed must he have in order to clear the canyon?
Assuming the motorcycle acts like a point mass (we don't want him to make it across with just his front wheel!) just drag out your good old equations for a trajectory:
for initial velocity v at angle t, the range is
r = (v^2 sin 2t)/g
10 = (v^2 sin 30°)/9.8
v^2 = 20 * 9.8 = 196
v = 14m/s or 50.4 km/hr
Just to check:
he'll be in the air 10/14 = .71 sec
his vertical velocity starts at 14 sin 15° = 3.62m/s
v = 3.62 - 9.8t
when v=0 (top of the arc) t = .36 sec
just about right. He has to go up and then come back down, taking .72 sec in all.