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hiralpatel121

  • 3 years ago

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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  1. JopHP
    • 3 years ago
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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

  2. ghazi
    • 3 years ago
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    @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...

  3. JopHP
    • 3 years ago
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    i know that @ghazi but those are the permitted givens

  4. rkphysics
    • 3 years ago
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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

  5. CarlosGP
    • 3 years ago
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    Here goes the solution. Potential Energy turns into Kinetic and Rotation Energy

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