Question 117

A bouncing tennis ball is dropped from a height of 32 metre. The ball rebounds each time to a height equal to half the height of the previous bounce. The approximate distance travelled by the ball when it hits the ground for the eleventh time, is:

Solution

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Total distance covered by the ball = $$32+(\dfrac{32}{2}+\dfrac{32}{2}+\dfrac{32}{2^2}+\dfrac{32}{2^2}+\dfrac{32}{2^3}+...)$$

=$$32+2(\dfrac{32}{2}+\dfrac{32}{2^2}+\dfrac{32}{2^3}+...)$$

$$\Rightarrow$$ $$32 + 64(\dfrac{1/2}{1-1/2}$$) = 96 m.

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