A ball is dropped from a height of 20 feet. It hits the ground and bounces up half as high as it fell. This is true for each successive bounce. What height does the ball reach on the fourth bounce

1.25 m

Step-by-step explanation:

Given that :

Initial height of drop = 20 meters

Rebound after each drop = 1/2 of height

Height after 4th rebound = 5th term of a geometric series

An = a1 * r^(n-1)

a1 = first term = initial drop height

r = common ratio = ratio of rebound per drop = 1/2 = 0.5

A5 = 20 * 0.5^(5 - 1)

A5 = 20 * 0.5^4

A5 = 20 * 0.0625

A5 = 1.25 m

Hence height after 4 reboubd = 1.25 m

Using another method :

1st bounce height = 20 * 0.5 = 10

2nd bounce height = 10 * 0.5 = 5

3rd bounce height = 5 * 0.5 = 2.5

4th bounce height = 2.5 * 0.5 = 1.25

49%

step-by-step explanation:

when you realize that you dont know just realize i just took a test and im so smart

"standard form:

3x2 + 14x − 10 = 0

solutions based on quadratic formula:

x1   = −14 − √ 142 − 4×3×(−10) /   2×3   = −14 − 2 × √ 79 /   6   ≈ −5.296

x2   = −14 + √ 142 − 4×3×(−10) /   2×3   = −14 + 2 × √ 79 /   6   ≈ 0.629