Question

A ball is thrown from a height of (156 feet) with an initial downward velocity of (8 ft/s). The ball's height (H) in feet after (T) seconds is given by (H = 156 - 8t - 16t²). How long after the ball is thrown does it hit the ground?

a) (2.5 seconds)

b) (3.0 seconds)

c) (3.5 seconds)

d) (4.0 seconds)

Answer 1

The ball hits the ground approximately 2.79 seconds after it is thrown.

Step-by-step explanation:

Given that the height of the ball (H) after t seconds is given by the equation H = 156 - 8t - 16t², we can find when the ball hits the ground by setting H equal to zero and solving for t.

0 = 156 - 8t - 16t²

Let's solve this quadratic equation using the quadratic formula.

t = (-b ± √(b² - 4ac)) / (2a)

Here, a = -16, b = -8, and c = 156. Plugging in these values, we get:

t = (-(-8) ± √((-8)² - 4(-16)(156))) / (2(-16))

t = (8 ± √(64 + 9984)) / (-32)

t = (8 ± √10048) / (-32)

t = (8 ± 100.24) / (-32)

Since time cannot be negative, we can discard the negative solution.

t = (8 + 100.24) / (-32)

t ≈ -2.79

Therefore, the ball hits the ground approximately 2.79 seconds after it is thrown. Since we are looking for the positive value, the correct answer is approximately 2.79 seconds.