A 20. kilogram object strikes the ground with 1960 joules of kinetic energy after falling freely from rest. How far above the ground was the object when it was released?

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Answer 1 · 2018-01-01T07:08:03

9.8 metres

The Kinetic energy of an object at a point of its motion is = #1/2mv^2#

Where,

        m = mass of the body

and v = instataneous velocity of the body at that point

So, According to the question,

#1/2(20)v^2 = 1960# [as m = 20 kg and K.E. = 1960 J]

#rArr 10v^2 = 1960#

#rArr v^2 = 196#

#rArr v = +-14#

As the direction of the velocity was same as the direction of the acceleration due to gravity, we will take the answer as positive.

Now, According to the Equations of motion,

#v = u + g * t# [Have to do it because the math compiler was mistakingly compiling to greater than sign]

#rArr 14 = 0 + 10 * t# [Body starts from rest and #g = 10 m s^-2#]

#rArr t = 1.4#

Now,

#h = 1/2g *t^2# [Man... its annoying]

#rArr h = 1/2 * 10 * (1.4)^2#

#rArr h = 5 * 1.96#

#rArr h = 9.8 # metres.