Question

A ball, dropped onto a hard floor from a height of 2 meters, bounces to a height of 1.8 meters, 90% of the initial height. Each subsequent bounce is 90% the height of the previous one. Find the total distance the ball travels up and down (i.e. after the first collision with the ground it has traveled 2 meters down and 1.8 meters up, giving a total distance of 3.8 meters).

Answer 1

Step 1)

As given A ball, dropped onto a hard floor from a height of 2 meters, bounces to a height of 1.8 meters, 90% of the initial height. Each subsequent bounce is 90% the height of the previous one we have to find the total distance the ball travels up and down

we can write,

after first collision with ground ball traveled : 2m down + 1.8m up = 2 + 1.8

90% of 1.8 = 1.62

after second collision with ground ball traveled : 1.8m down + 1.62m up = 1.8 + 1.62

90% of 1.62 = 1.458

after third collision with ground ball traveled : 1.62m down + 1.458m up = 1.62 + 1.458

90% of 1.458 = 1.3131

after fourth collision with ground ball traveled : 1.458m down + 1.3131m up = 1.458 + 1.3131

90% of 1.3131 = 1.18179

after fifth collision with ground ball traveled : 1.3131m down + 1.18179m up = 1.3131 + 1.8179

and so on,

Step 2)

Hence we can write total distance traveled by the ball is given by,

    --------------------------------------1)

we will evaluate,

we can see that,

Means given series is a geometric series with first first a = 1.8 and common ratio r = 0.9

we know that sum of the geometric series with first term a and common ration r is given by,

where r < 1

we have a = 1.8 and r = 0.9 < 1 hence we can say that sum of the given series is given by,

Hence we can write,

Put this value in equation 1) we can write,

Hence we can say that total distance traveled by the ball is 38 m