A ball is thrown vertically upward from the top of a building 128 feet tall with an initial velocity of 112 feet per second.
The distance (in feet) of the ball from the ground after seconds is
(a) After how many seconds does the ball strike the ground?
(b) After how many seconds will the ball pass the top of the building on its way down?
(a) After how many seconds does the ball strike the ground?
seconds

Ask by Allan Bates. in the United States

Jan 24,2025

Question

A ball is thrown vertically upward from the top of a building 128 feet tall with an initial velocity of 112 feet per second.
The distance (in feet) of the ball from the ground after seconds is
(a) After how many seconds does the ball strike the ground?
(b) After how many seconds will the ball pass the top of the building on its way down?
(a) After how many seconds does the ball strike the ground?
seconds

Ask by Allan Bates. in the United States

Jan 24,2025

UpStudy AI · Accepted Answer

To determine when the ball strikes the ground, we need to find the time $$ t $$ when the distance $$ s(t) $$ from the ground is zero.

Given the equation:
$$ s(t) = 128 + 112t - 16t^2 $$

Set $$ s(t) = 0 $$ and solve for $$ t $$:
$$ 0 = 128 + 112t - 16t^2 $$
$$ -16t^2 + 112t + 128 = 0 $$

To simplify, divide the entire equation by -16:
$$ t^2 - 7t - 8 = 0 $$

Now, solve the quadratic equation using the quadratic formula:
$$ t = \frac{7 \pm \sqrt{(-7)^2 - 4(1)(-8)}}{2(1)} $$
$$ t = \frac{7 \pm \sqrt{49 + 32}}{2} $$
$$ t = \frac{7 \pm \sqrt{81}}{2} $$
$$ t = \frac{7 \pm 9}{2} $$

This gives two potential solutions:
$$ t = \frac{16}{2} = 8 $$
$$ t = \frac{-2}{2} = -1 $$

Since time cannot be negative, we discard $$ t = -1 $$ and accept:
$$ t = 8 \text{ seconds} $$

**Answer:**

Problem a Answer:

The ball strikes the ground after $$ 8 $$ seconds.
The ball strikes the ground after 8 seconds.

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