Suppose that a ball is thrown from the top of a 64 -foot building. The function models the height in feet, , of the ball seconds after it is thrown.
a) Find the zero(s) of the function. 〔You may use fractions or decimals. Use a comma to separate values, If necessary)

The zero(s) of the function:
b) If we assume that ground level means , when dees the ball hit the ground?

The ball hits the ground after 4
seconds
c) Find the time(s) the ball will be 84 feet above ground. {You may use fractions or decimals. Use a comma to separate values, if necessary}

The ball is 84 feet above ground after 100
feet
d) The ball, will be at its highest point after seconds. Find the height of the ball at that time.

The height of the ball after 1.5 seconds is Select an answer

Ask by Warner Guerrero. in the United States

Mar 31,2025

Question

Suppose that a ball is thrown from the top of a 64 -foot building. The function models the height in feet, , of the ball seconds after it is thrown.
a) Find the zero(s) of the function. 〔You may use fractions or decimals. Use a comma to separate values, If necessary)

The zero(s) of the function:
b) If we assume that ground level means , when dees the ball hit the ground?

The ball hits the ground after 4
seconds
c) Find the time(s) the ball will be 84 feet above ground. {You may use fractions or decimals. Use a comma to separate values, if necessary}

The ball is 84 feet above ground after 100
feet
d) The ball, will be at its highest point after seconds. Find the height of the ball at that time.

The height of the ball after 1.5 seconds is Select an answer

Ask by Warner Guerrero. in the United States

Mar 31,2025

UpStudy AI · Accepted Answer

**a)** To find the zeros of the function
$$
h(t) = -16t^2 + 48t + 64,
$$
set $$ h(t) = 0 $$:
$$
-16t^2 + 48t + 64 = 0.
$$
Multiply both sides by $$-1$$ to simplify:
$$
16t^2 - 48t - 64 = 0.
$$
Divide the entire equation by $$16$$:
$$
t^2 - 3t - 4 = 0.
$$
Factor the quadratic:
$$
(t - 4)(t + 1) = 0.
$$
Thus, the solutions are
$$
t = 4 \quad \text{and} \quad t = -1.
$$
Since time cannot be negative in this context, the meaningful zero is at $$ t = 4 $$ seconds. However, listing the zeros as requested:
$$
-1,\ 4.
$$

**b)** When the ball hits the ground, $$ h(t) = 0 $$. From part (a), this occurs at $$ t = 4 $$ seconds (ignoring the negative solution). Therefore, the ball hits the ground at
$$
t = 4 \text{ seconds}.
$$

**c)** To find the time(s) when the ball is $$ 84 $$ feet above ground, set $$ h(t) = 84 $$:
$$
-16t^2 + 48t + 64 = 84.
$$
Subtract $$84$$ from both sides:
$$
-16t^2 + 48t + 64 - 84 = 0 \quad \Longrightarrow \quad -16t^2 + 48t - 20 = 0.
$$
Multiply through by $$-1$$:
$$
16t^2 - 48t + 20 = 0.
$$
Divide the entire equation by $$4$$:
$$
4t^2 - 12t + 5 = 0.
$$
Use the quadratic formula:
$$
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
with $$ a = 4,\ b = -12,\ c = 5 $$:
$$
t = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(4)(5)}}{2(4)} = \frac{12 \pm \sqrt{144 - 80}}{8} = \frac{12 \pm \sqrt{64}}{8}.
$$
Since $$ \sqrt{64} = 8 $$, we have:
$$
t = \frac{12 \pm 8}{8}.
$$
This gives two solutions:
$$
t = \frac{12 + 8}{8} = \frac{20}{8} = 2.5 \quad \text{and} \quad t = \frac{12 - 8}{8} = \frac{4}{8} = 0.5.
$$
Thus, the ball is $$84$$ feet above ground at
$$
t = 0.5 \text{ seconds and } t = 2.5 \text{ seconds}.
$$

**d)** The ball reaches its highest point at $$ t = 1.5 $$ seconds. Substitute $$ t = 1.5 $$ into $$ h(t) $$:
$$
h(1.5) = -16(1.5)^2 + 48(1.5) + 64.
$$
First, calculate $$ (1.5)^2 $$:
$$
(1.5)^2 = 2.25.
$$
Then,
$$
-16(2.25) = -36,\quad 48(1.5) = 72.
$$
So,
$$
h(1.5) = -36 + 72 + 64 = 100.
$$
The height of the ball at $$ t = 1.5 $$ seconds is
$$
100 \text{ feet}.
$$
**a)** The zeros of the function are $$ -1 $$ and $$ 4 $$. **b)** The ball hits the ground after $$ 4 $$ seconds. **c)** The ball is $$ 84 $$ feet above ground at $$ 0.5 $$ and $$ 2.5 $$ seconds. **d)** The height of the ball at $$ 1.5 $$ seconds is $$ 100 $$ feet.

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