A ball is dropped from 248 feet above the ground

Let us say

The Height from which the ball was first dropped as H0 = 248 feet
The Height reached by the ball after second bounce is H1
The Height reached by the ball after second bounce is H2 = 62 feet

Given , Height the ball reaches after bounce is a fraction of the height the ball reached in the previous bounce

The Height reached by the ball after second bounce = H2 = X of H1
==> 62 = X * H1 ...................Equation 1

The Height reached by the ball after first bounce = H1 = X of H0

==> H1 = X * 248 ...................Equation 2

Substituting value of H1 from Equation 2 to Equation 1

62 = X * H1 = X * X * 248

X^2 = 62/248 = 31/124 = 1/4

X = 1/2

Answer is A. 1/2


This can be solved by Geometric Progression formula as well

Term n = Term 1 * R ^ (n -1)

Here

R is nothing but the fraction, the question refers to

Term 1 is the Height from which the ball is dropped from = 248 feet
Term 2 is the Height reached by the ball after first bounce
.
.
.
Term n can be Term 3, which is the height the ball reached after the second bounce = 62 feet

Putting the above in equation

62 = 248 * x ^ (3-2)

62 = 248 * X ^ 2

62/248 = X ^ 2

X ^ 2 = 1/4

X = 1/2