Question

1. a model rocket is fired vertically upward from the ground. the distance (in feet) that the rocket travels from the ground after t seconds is given by (s(t)=-16t^{2}+560t). (i) find the speed of the rocket 3 sec after being fired. is it going up or down? (ii) how high does the rocket get?

Explanation:

Step1: Find the velocity function

The velocity function $v(t)$ is the derivative of the position - function $s(t)$. Given $s(t)=-16t^{2}+560t$, using the power rule $\frac{d}{dt}(at^{n}) = nat^{n - 1}$, we have $v(t)=s^\prime(t)=-32t + 560$.

Step2: Calculate the velocity at $t = 3$

Substitute $t = 3$ into $v(t)$: $v(3)=-32\times3 + 560=-96 + 560 = 464$ feet per second. Since $v(3)>0$, the rocket is going up.

Step3: Find the time when the rocket reaches its maximum height

At the maximum - height, the velocity of the rocket is $0$. Set $v(t)=0$, so $-32t + 560 = 0$. Solving for $t$:
\[

$$\begin{align*} -32t&=-560\\ t&=\frac{560}{32}=\frac{35}{2}=17.5 \end{align*}$$

\]

Step4: Calculate the maximum height

Substitute $t = 17.5$ into the position function $s(t)$: $s(17.5)=-16\times(17.5)^{2}+560\times17.5=-16\times306.25 + 9800=-4900+9800 = 4900$ feet.

Answer:

(i) The speed of the rocket 3 seconds after being fired is 464 feet per second and it is going up.
(ii) The rocket reaches a height of 4900 feet.