Question

a. A rocket of mass $m$is launched straight up with thrust ${\overrightarrow{F}}_{thrust}$. Find an expression for the rocket’s speed at height $h$if air resistance is neglected.

b. The motor of a $350$g model rocket generates$9.5$ N thrust. If air resistance can be neglected, what will be the rocket’s speed as it reaches a height of $85$ m?

Short answer

a). An expression for the rocket's speed at height $h$if the air is neglected localid="1649520749864" $v=\sqrt{2\left(\frac{F_{thrust}-mg}{m}\right)h}$

b). The speed of the rocket at height $85$m is 54 m/s

Step-by-step solution

Part a). Step 1. Finding an expression

We can start solution with Newton's second law

$$\begin{gathered} ma=F_{thrust}-mg \\ a=\frac{F_{thrust}-mg}{m} \end{gathered}$$

Now, we can determine speed from an expression that includes acceleration and height

localid="1649520770861" $$\begin{gathered} v=\sqrt{2ah} \\ v=\left(\sqrt{2·\left(\frac{F_{thrust}-mg}{m}\right)·h}\right) \end{gathered}$$

Part b) Step 1. Find the speed

Given that,$$\begin{gathered} m=0.35kg \\ {F}_{thrust}=9.5N \\ h=85m \\ g=9.81m/s^2 \end{gathered}$$

If we substitute values in the previous rocket's speed expression, we get

$v=\sqrt{2\times \frac{(9.5-0.35\times 9.81)}{0.35}\times 85}=54.28\simeq 54m/sapprox.$