a ball is dropped on the ground.The time interval between the 1st and 2nd bounce is 1.2s.The interval between the 1st and 2nd and 3rd bounce is 0.9s.Assume that successive time intevals form a geometric progression with an infinite number of terms.Hence find the total time for which the ball bounces
So we are to assume that the common ratio is .9/1.2 sec or 3/4 second
our total time = 1.2 + 1.2(3/4) + 1.2(3/4)^2 + ...
sum(all terms) = a/(1-r)
= 1.2/(1-3/4)
= 1.2/(1/4)
= 4.8 seconds
I think the answer is correct
I u sure is correct
What if , we say,I don't know,I think what we're asked is to find the sum to infinity of the time,which rep the terms as main :ar=1.2 ,ar²=0.9, solving it, we get r=0.75 and a=1.6,we solve with S, and get 6.4 ,how about that ?
To find the total time for which the ball bounces, we need to determine the sum of an infinite geometric series.
Let's break down the problem step by step:
Step 1: Identify the common ratio (r) of the geometric progression.
In this case, we know that the time interval between each bounce forms a geometric progression. Let's denote the first time interval as a and the common ratio as r.
Given:
First time interval = a = 1.2 seconds
Second time interval = a * r = 0.9 seconds
To find the common ratio, divide the second time interval by the first time interval:
a * r / a = 0.9 / 1.2 = 3/4
So, the common ratio (r) is 3/4.
Step 2: Calculate the sum of an infinite geometric series.
The sum formula for an infinite geometric series is given by:
S = a / (1 - r),
where S is the sum, a is the first term, and r is the common ratio.
In this case, S represents the total time for which the ball bounces, a represents the first time interval (1.2 seconds), and r represents the common ratio (3/4).
Substituting these values into the formula, we have:
S = (1.2) / (1 - 3/4)
Step 3: Simplify the expression.
To simplify the expression, we need to find a common denominator:
S = (1.2) / (4/4 - 3/4)
= (1.2) / (1/4)
= 1.2 * (4/1)
= 4.8
Hence, the total time for which the ball bounces is 4.8 seconds.