A ball is dropped from 248 feet above the ground level and after the second
bounce it rises to the height of 62 feet.
If the height to which the ball rises
after each bounce is always the same
fraction of the height reached on its
previous bounce, what is this
fraction?

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

395.84

step-by-step explanation:

2πrh+2πr²

2 (pie)(radius)(height)+2(pie)(radius)^2

answer: a

step-by-step explanation: the answer is a due to the rule of symmetry; that a line of symmetry must divide a figure into 2 pieces that are the exact same, only reflected over a line of symmetry. b and c do not have lines of symmetry and would not divide the figure evenly, thus making the answer a.