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A 11.7 ft wide ditch with the approach roads at an angle of 15° with the horizontal. With what minimum speed should a motorbike be moving on the road so that it safely crosses the ditch?


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The center of mass of motor bike is C and it is in the shown position just before it is supposed to have taken off.  C' is the position when the bike lands safely ie., the rear part B must land on the road on the other side.

CC' is the range for the flight of the bike.  CC' = R
                   = 11.7' + 2.5' cos 15° + 2.5' Cos 15° = 16.53'

We assume that the angle of projection of center of mass is 15°.

Range for a two dimensional projectile : R = u² Sin 2Ф / g  = u² Sin (2*15°) / g
          u = √(2 R g ) = √(2*16.53 * 9.8 / 0.3048) = 32.6 feet / sec

We do not consider some other factors like rotation about center of mass, friction etc. while answering this question.

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