Explain why a tennis ball bounces higher on hills than in plains?
(A) \[g\] on hills is less
(B) \[g\] on hills is more
(C) \[g\] on hills is same as plains
(D) Independent of \[g\]

Hint: The height to which an object bounces at a particular place depends upon the force acting on the body towards the surface at that place. When a body bounces or rises, the weight of the object, acting normally downwards tends to bring the body down. We know this phenomenon as gravity.

Formula Used: \[g'=g{{(1+\dfrac{h}{R})}^{-2}}\]

Complete step by step answer:
From the hint, we can infer that more the force acting on the body, lesser will be the bounce.
The force acting on a body towards the earth is its weight which is a product of the mass of the body and the acceleration due to gravity at that point. The value of acceleration due to gravity \[g\] is maximum at the earth’s surface and varies as we move up or down from the surface. As we start moving upwards from the earth’s surface, acceleration due to gravity varies as \[g'=g{{(1+\dfrac{h}{R})}^{-2}}\] where \[h\] is the height above the earth’s surface.
We’ve successfully established the fact that the value of \[g\] is more at the equator (or plains) than at a hill. Hence we can also say that the weight experienced by the body at the equator will be more than the weight experienced at the hill. Hence the ball will bounce higher at the hill.
So we can now say that the reason a tennis ball bounces higher on hills than in plains is that \[g\] on hills is less.
Hence, option (A) is correct.

Note: Apart from the variation with height, \[g\] also varies with the radius of the earth and the angular speed of rotation. Hence, \[g\] is slightly larger at the poles than at the equator. Hence, a tennis ball will bounce more at the equator than at poles. However, if the earth stopped rotating, the value of \[g\] will become the same at all places on the earth’s surface.