ball is dropped from a height of 10 feet and bounces. Each bounce is 3/4 of the height
of the bounce before. Thus, after the ball hits the floor for the first time, the ball rises to
a height of 10(3/4) = 7.5 feet, and after it hits the second floor for the second time, it
rises to a height of 7.5(3/4) = 10(3/4)² = 5.625 feet. (Assume g = 32 ft/sec² and that
there is no air resistance.)
a. Find an expression for the height to which the ball rises after it hits the floor for the
nth time.
b. Find an expression for the total vertical distance the ball has travelled when it hits
the floor for the first, second, third, and fourth times.
c. Find an expression for the total vertical distance the ball has travelled when it hits
the floor for the nth time
a. To find an expression for the height to which the ball rises after it hits the floor for the nth time, we can use the pattern observed in the problem:
After the first bounce, the ball rises to a height of 10(3/4) = 7.5 feet.
After the second bounce, the ball rises to a height of 7.5(3/4) = 5.625 feet.
After the third bounce, the ball rises to a height of 5.625(3/4) = 4.21875 feet.
We can see that for each bounce, the height is multiplied by 3/4. We can use this pattern to find an expression for the height after the nth bounce.
Let H(n) represent the height the ball reaches after the nth bounce. We can express H(n) as:
H(n) = 10(3/4)^(n-1)
b. To find an expression for the total vertical distance the ball has travelled when it hits the floor for the first, second, third, and fourth times, we need to consider both the distance covered during the upward motion and the downward motion.
During the upward motion, the ball covers the same distance as the height it reaches after each bounce. We can use the expression derived in part a to represent this distance for each bounce.
During the downward motion, the ball covers the same distance as the height from which it was dropped. In this case, it is 10 feet.
To find the total distance traveled, we multiply the total number of bounces (n) by the distance covered during the upward motion and add the distance covered during the downward motion.
Total distance for the first bounce = H(1) + 10
Total distance for the second bounce = H(2) + H(1) + 10
Total distance for the third bounce = H(3) + H(2) + H(1) + 10
Total distance for the fourth bounce = H(4) + H(3) + H(2) + H(1) + 10
We can generalize this expression for the nth bounce:
Total distance for the nth bounce = H(1) + H(2) + ... + H(n) + 10
c. To find an expression for the total vertical distance the ball has traveled when it hits the floor for the nth time, we can simplify the expression found in part b using the geometric series formula.
The sum of a geometric series can be calculated using the formula:
S = a(1 - r^n) / (1 - r)
In this case, a = H(1) and r = 3/4. We can substitute these values into the formula to find the expression for the total distance.
Total distance for the nth bounce = H(1) * (1 - (3/4)^n) / (1 - 3/4) + 10
Simplifying further:
Total distance for the nth bounce = H(1) * (4/1) * (1 - (3/4)^n) + 10
Total distance for the nth bounce = 40 * (1 - (3/4)^n) + 10
Therefore, the expression for the total vertical distance the ball has traveled when it hits the floor for the nth time is 40 * (1 - (3/4)^n) + 10.
(a) really? After all that introduction, you don't know that
h(n) = 10(3/4)^n
(b) as with all geometric series, Sn = a(1-r^n)/(1-r)
In this case, we double every term but the first, so
S5 = 2*10(1 - (3/4)^4)/(1 - 3/4) - 10
(c) see (b)