In figure i shows 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 road so that it crosses the ditch safely ? Assume that the length of bike is 5 ft and it leaves the road when he front part runs out of the approach road.
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Please see diagram enclosed.
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
This is a simple answer, we do not consider some other factors like rotation about center of mass, friction etc.
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
This is a simple answer, we do not consider some other factors like rotation about center of mass, friction etc.
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Anonymous:
Thank you
thanx n u r welcom
your welcome
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