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A solid sphere rolls down a hill of height 60 m. What is the velocity of the ball when it reaches the bottom?

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since it rolls down from a height so it's initial velocity U=0 & it's Acc.=the gravititional Acc=9.8 m/sec^2 therefore by applying the law of motion V^2=U^2+2GS therefore: V^2=0+2(9.8)(60)=1176 m^2/sec^2 So. V=34.29 m/sec
@JopHP what you've mentioned here is case of free fall when angle of inclination is 90 degrees in this case ball is rolling down the hill so velocity will be different @hiralpatel can you mention the angle of inclination...but there is one more question angle of inclination will keep changing...
i know that @ghazi but those are the permitted givens

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\[\[V_{final}^2= 2[g.\cos \theta ]*[h/\cos \theta]\] for any inclination theta, so final velocity will be same if it is a steep hill or a nearly flat one!!! assuming the ball started from rest and neglecting friction
Here goes the solution. Potential Energy turns into Kinetic and Rotation Energy

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