Question

For an object, dropped from a height of 16 m, the ratio of its kinetic energy and potential energy above the ground at half the height is

• A. 1 : 2
• B. 2 : 1
• C. 1 : 4
• D. 1 : 1

Answer 1

The correct option is D 1 : 1

At half the height, the ball will be 8m away from the ground, then at point 'B' $\Rightarrow P.E.=mgh=8mg$
From third equation of motion we have,$v^2-u^2=2as,s=8\text{u=0, since object is dropped from rest}{v}^2=2g\times 8v^2=16g\text{K.E. is given as}K.E=\frac{1}{2}m{v}^2\text{Substituting the value of v in K.E., we get}K.E.=\frac{1}{2}\times m\times 16g=8mg\text{So ratio of}\text{potential and kinetic energy is}K.E:P.E=1:1$