Question

a model rocket is launched with an initial upward velocity of 195 ft/s. the rocket’s height h (in feet) after t seconds is given by the following. h = 195t - 16t². find all values of t for which the rocket’s height is 87 feet. round your answer(s) to the nearest hundredth. (if there is more than one answer, use the “or” button.)

Explanation:

Step1: Set up the equation

Set $h = 87$ in the height - time equation $h=195t - 16t^{2}$. So we get $87=195t - 16t^{2}$. Rearrange it to the standard quadratic - form $ax^{2}+bx + c = 0$. We have $16t^{2}-195t + 87 = 0$.

Step2: Identify the coefficients

For the quadratic equation $16t^{2}-195t + 87 = 0$, $a = 16$, $b=-195$, and $c = 87$.

Step3: Use the quadratic formula

The quadratic formula is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substitute $a = 16$, $b=-195$, and $c = 87$ into the formula:
\[

$$\begin{align*} t&=\frac{-(-195)\pm\sqrt{(-195)^{2}-4\times16\times87}}{2\times16}\\ &=\frac{195\pm\sqrt{38025-5568}}{32}\\ &=\frac{195\pm\sqrt{32457}}{32}\\ &=\frac{195\pm180.16}{32} \end{align*}$$

\]

Step4: Calculate the two values of t

For the plus - sign: $t=\frac{195 + 180.16}{32}=\frac{375.16}{32}\approx11.72$.
For the minus - sign: $t=\frac{195 - 180.16}{32}=\frac{14.84}{32}\approx0.46$.

Answer:

$t\approx0.46$ or $t\approx11.72$