An automobile moving at a speed of 72 km/h reaches the foot of a smooth incline when the engine is switched off.How much distance does the automobile go up the incline before stopping? The incline makes an angle of 30° with the horizontal.Take g =9.8
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conversion 72*5/18 is 20m/s
now here the whole kinetic energy will be converted into potential energy when the automobile stops thus
1/2mv^2=mgh
20*20/2=9.8*h
200/9.8=h
20.4=h
so the body will go 20.4 metres high vertically but now here the inclined angle given is 30° now vertical creates 90° hence 90-30 gives 60° thus cos60° will be multiplied now we get 20.4*1/2 thus approximately 10 meters
now here the whole kinetic energy will be converted into potential energy when the automobile stops thus
1/2mv^2=mgh
20*20/2=9.8*h
200/9.8=h
20.4=h
so the body will go 20.4 metres high vertically but now here the inclined angle given is 30° now vertical creates 90° hence 90-30 gives 60° thus cos60° will be multiplied now we get 20.4*1/2 thus approximately 10 meters
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