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 32 ft/sec is given by the equation h = 32t - 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 2 seconds.
(simplify your answer. type an integer or a decimal. round to the nearest tenth.)
b) how long after it is thrown upward will the object be 12 ft above the point of release?
a. the object will reach a height of 12 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 12 ft at second(s) after its release. the object will also reach a height of 12 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 12 ft above the point of release.

Explanation:

Step1: Set up the equation for part a

When the object returns to the original height, $h = 0$. So we set $32t-16t^{2}=0$.

Step2: Factor the equation

Factor out $16t$: $16t(2 - t)=0$.

Step3: Solve for $t$

Using the zero - product property, either $16t=0$ or $2 - t = 0$. $t = 0$ is the initial time of release. For $2 - t=0$, we get $t = 2$.

Step4: Set up the equation for part b

We want to find $t$ when $h = 12$. So the equation is $32t-16t^{2}=12$. Rearrange it to the standard quadratic form $16t^{2}-32t + 12=0$. Divide through by 4 to simplify: $4t^{2}-8t + 3=0$.

Step5: Solve the quadratic equation

Use the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here $a = 4$, $b=-8$, $c = 3$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-8)^{2}-4\times4\times3=64 - 48 = 16$. Then $t=\frac{8\pm\sqrt{16}}{8}=\frac{8\pm4}{8}$.

Step6: Find the two values of $t$

For the plus - sign: $t=\frac{8 + 4}{8}=\frac{12}{8}=1.5$. For the minus - sign: $t=\frac{8 - 4}{8}=\frac{4}{8}=0.5$.

Answer:

a) 2
b) B. The object will first reach a height of 12 ft at 0.5 second(s) after its release. The object will also reach a height of 12 ft at 1.5 second(s) after its release.