quantumliner
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
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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
Hey, thanks! I understood it now.
your clear explanation makes it so much simpler. much appreciated.