the height h (in feet) relative to the point of release of an object t sec after it is thrown straight upward with an initial velocity of 64 ft/sec is given by the equation h = 64t - 16t².
a) how long after it is thrown upwards will it take the object to return to the original height at which is was released?
the object will return in 4 seconds.
b) how long after it is thrown upward will the object be 32 ft above the point of release?
(simplify your answer. type an integer or a decimal. round to the nearest tenth.)
a. the object will reach a height of 32 ft above the point of release once at second(s) after its release. (simplify your answer. type an integer or a decimal. round to the nearest tenth.)
b. the object will first reach a height of 32 ft at second(s) after its release. the object will also reach a height of 32 ft at second(s) after its release. (simplify your answer. type an integer or a decimal. round to the nearest tenth.)
c. the object will never reach a height 32 ft above the point of release.

Question

the height h (in feet) relative to the point of release of an object t sec after it is thrown straight upward with an initial velocity of 64 ft/sec is given by the equation h = 64t - 16t².
 a) how long after it is thrown upwards will it take the object to return to the original height at which is was released?
 the object will return in 4 seconds.
 b) how long after it is thrown upward will the object be 32 ft above the point of release?
 (simplify your answer. type an integer or a decimal. round to the nearest tenth.)
 a. the object will reach a height of 32 ft above the point of release once at second(s) after its release. (simplify your answer. type an integer or a decimal. round to the nearest tenth.)
 b. the object will first reach a height of 32 ft at second(s) after its release. the object will also reach a height of 32 ft at second(s) after its release. (simplify your answer. type an integer or a decimal. round to the nearest tenth.)
 c. the object will never reach a height 32 ft above the point of release.

Accepted Answer

# Explanation:
## Step1: Solve for part a)
Set $h = 0$ in the equation $h=64t - 16t^{2}$.
$$
\begin{align*}
0&=64t-16t^{2}\
16t^{2}-64t&=0\
16t(t - 4)&=0
\end{align*}
$$
Using the zero - product property $ab = 0$ implies $a=0$ or $b = 0$. Here $16t=0$ gives $t = 0$ (initial time) and $t - 4=0$ gives $t = 4$.

## Step2: Solve for part b)
Set $h = 32$ in the equation $h=64t - 16t^{2}$.
$$
\begin{align*}
32&=64t-16t^{2}\
16t^{2}-64t + 32&=0\
t^{2}-4t + 2&=0
\end{align*}
$$
Use the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c=0$. Here $a = 1$, $b=-4$, $c = 2$.
$$
\begin{align*}
t&=\frac{4\pm\sqrt{(-4)^{2}-4\times1\times2}}{2\times1}\
&=\frac{4\pm\sqrt{16 - 8}}{2}\
&=\frac{4\pm\sqrt{8}}{2}\
&=\frac{4\pm2\sqrt{2}}{2}\
&=2\pm\sqrt{2}
\end{align*}
$$
$\sqrt{2}\approx1.414$. So $t_1=2+\sqrt{2}\approx2 + 1.414=3.4$ and $t_2=2-\sqrt{2}\approx2-1.414 = 0.6$

# Answer:
a) $4$ seconds.
b) The object will be $32$ ft above the point of release at approximately $0.6$ seconds (first time) and $3.4$ seconds (second time). So A. The object will reach a height of $32$ ft above the point of release once at $0.6$ seconds after its release. B. The object will first reach a height of $32$ ft at $0.6$ second(s) after its release. The object will also reach a height of $32$ ft at $3.4$ second(s) after its release.

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QUESTION IMAGE

Question

Explanation:

Step1: Solve for part a)

Step2: Solve for part b)

Answer: