Question

A model rocket is released straight upward with an initial speed of 60.5 m/s. It accelerates with a 

constant upward acceleration of 2.50 m/s2 until its engines stop at an altitude of 175. m. (a) What can you 

say about the motion of the rocket after its engines stop? (b) What is the maximum height reached by the 

rocket? (c) How long after liftoff does the rocket reach its maximum height? (d) How long is the rocket in 

the air?

Answer 1

a) After the engines stop the motion of the rocket is free fall.

b) The height of the rocket after engines stop is given by the kinematic law:

$$h = h_0+v_1t-\dfrac{gt^2}{2}$$

where $h_0 = 175m$ is the initial height, $v_1$ is the speed after engines stop, $g = 9.81m/s^2$ is the gravitational acceleration, and $t$ is time.

The height reached by the rocket just before engine stops is given by the same law, but with different acceleration:

$$h_0= v_0t_1+\dfrac{at_1^2}{2}$$

where $v_0 = 60.5m/s$, and $a = 2.5m/s^2$, $t_1$ is the moment of time when engines stop. Solving the quadratic equation, obtain:

$$1.25t_1^2 + 60.5t_1-175 = 0\\ t_1 \approx 2.74s$$

The final speed reached after the first part of the motion is:

$$v_1 = v_0 + at_1$$

Accoridng to the energy conservation law, the kinetic energy just after engine stop is equal to the potential energy at maximum height $h_{max}$ :

$$\dfrac{mv_1^2}{2} = mgh_{max}\\ h_{max} = \dfrac{v_1^2}{2g} = \dfrac{(v_0+at_1)^2}{2g}\\ h_{max} = \dfrac{(60.5+2.5\cdot 2.74)^2}{2\cdot 9.81} \approx 231.2m$$

c) Substituting $h = h_{max}$ and expression for $v_1$ into the first equation, find the time required to reach the maximum height (counting from $t_1$):

$$h_{max} = h_0+(v_0+at_1)t_{2}-\dfrac{gt_{2}^2}{2}$$

Solving the quadratic equation, find:

$$4.905t_2^2-67.344t_2+56.2=0\\ t_2 = 12.84s$$

Counting from the liftoff:

$$t_{max} = t_1 + t_2 = 15.58s$$

d) Considering the motion from the maximum point, the heigth of the rocket is given as:

$$x = h_{max}-\dfrac{gt^2}{2}$$

It lands when $x = 0$. thus, it takes time :

$$h_{max}-\dfrac{gt_3^2}{2} = 0\\ t_3 =\sqrt{\dfrac{2h_{max}}{g}}\\ t_3 \approx 6.87s$$

Thus, the total time in air is:

$$t_{air} = t_{max} + t_3 = 22.45s$$

Answer. a) free fall with initial velocity, b) 231.2 m, c) 15.8s, d) 22.45s.