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

Step-by-step 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)=(V_(fa) - V_(oa))/(a)` (6)

`t_(a)=(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)=-(V_(ob)^(2))/(2g)` (10)

`H_(max)=-((60.07 m/s)^(2))/(2(-9.8 m/s^(2)))` (11)

Finally:

`H_(max)=184.10 m`