Question

A ball is dropped from a height of 6 feet and begins bouncing.the height of each bounce is three-fourths the height of the previous bounce.find the total vertical distance travelled by the ball.

Answer 1

Answer:

42 feet

Step-by-step explanation:

so when the ball hit the ground for the 1st time it travels a distance of 6 feet.

for later bounces let Di be the distance traveled  up and down.

D1= 6 feet

D2 and D3 would be

$D2 =up+ down\\=6(3/4)+6(3/4)\\=12(3/4)$

similarly D3 is

$D3=6(3/4)(3/4)+6(3/4)(3/4)\\=12(3/4)^2$

by repeating this process we can find the vertical distance traveled by the ball

$D=6+12(3/4)+12(3/4)^2+......$

=6+12∑ $(3/4)^(n+1)$

where n=0,1,2,.....∞

=6+12(3/4)∑$(3/4)^n$

$=6+9[1/(1-(3/4)]$

$=6+9(4)$

$=42 feet$

Answer 2

Given:
ball is dropped from a height of 6 feet. 
each bounce has a height of 3/4 the height of the previous bounce.

1st bounce: 6 x 3/4 = 4.5 feet
2nd bounce: 4.5 x 3/4 = 3.375 feet
3rd bounce: 3.375 x 3/4 = 2.531
4th bounce: 2.531 x 3/4 = 1.898
5th bounce: 1.898 x 3/4 = 1.424

The equation for the height of the bounce is:
3/4 = 0.75

h(n) = 6(0.75)^n ; where n is the number of the bounce.

Total vertical distance traveled by the ball means the up and down the ball traveled. So, we add up the computed distance and multiply it by 2.

4.5 + 3.375 + 2.531 + 1.898 + 1.424 = 13.728
13.728 x 2 = 27.456 feet

The ball had a total vertical distance of 27.456 feet after bouncing 5 times.