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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 deg...Asked by Susan
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?
Answers
Answered by
Steve
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.
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.
Answered by
Eddie
14m/s
Answered by
Ibrahim
Yes
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