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?

1 Answer

9.8 metres

Explanation:

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.