Question

A model rocket is launched straight upward with an initial speed of 52.1 m/s. It acceler- ates with a constant upward acceleration of 2.63 m/s2 until its engines stop at an altitude of 170 m. What is the maximum height reached by the rocket?

Answer 1

Answer: 184.10 m

Explanation:

In order to solve this problem, we have to divide it into two parts:

a) When the rocket is moving with an acceleration $a=2.63 m/s^{2}$

b) When the rocket is moving with an acceleration $a=g=-9.8 m/s^{2}$

So, for part a we have the following data:

$V_{oa}=52.1 m/s$ is the initial speed from ground

$a=2.63 m/s^{2}$ is the upward acceleration due to the rocket's engines

$h_{a}=170 m$ is the height the rocket reaches with this acceleration

We have to find the "final speed" $V_{fa}$ the rocket has when it reaches $h_{a}$:

$V_{fa}^{2}=V_{oa}^{2}+2ah_{a}$ (1)

$V_{fa}=\sqrt{V_{oa}^{2}+2ah_{a}}$ (2)

$V_{fa}=\sqrt{(52.1 m/s)^{2}+2(2.63 m/s^{2})(170 m)}$ (3)

$V_{fa}=60.07 m/s$ (4)

Then we have to find the time it took to the rocket to reach this velocity $V_{fa}$ with the following equation:

$V_{fa}=V_{oa} + a t_{a}$ (5)

$t_{a}=\frac{V_{fa} - V_{oa}}{a}$ (6)

$t_{a}=\frac{60.07 m/s - 52.1 m/s}{2.63 m/s^{2}}$ (7)

$t_{a}=3.03 s$ (8)

Now we move to part b:

At this point the acceleration is the acceleration due gravity ($a=g=-9.8 m/s^{2}$) and the final velocity $V_{fa}$ we calculated in equation (4) part a, is now the initial velocity $V_{ob}$ in this part b:

$V_{ob}=V_{fa}=60.07 m/s$

On the other hand, it is known that in projectile motion (as this situation) the maximum height $H_{max}$ is when the velocity of the rocket is zero ($V_{fb}=0$).

So, we will use equation (1) again but with this new data:

$V_{fb}^{2}=V_{ob}^{2}+2gH_{max}$ (9)

Isolating $H_{max}$:

$H_{max}=-\frac{V_{ob}^{2}}{2g}$ (10)

$H_{max}=-\frac{(60.07 m/s)^{2}}{2(-9.8 m/s^{2})}$ (11)

Finally:

$H_{max}=184.10 m$