Chapter 11, Problem 61A

To determine

The additional height by which the rocket will rise.

Answer to Problem 61A

  $20\text{m}$

Explanation of Solution

Given:

The mass of test rocket, $m=10.0\text{kg}$

The rocket’s kinetic energy when the engine stops firing, $KE=1960\text{J}$

Formula used:

The kinetic energy of an object of mass $m$ travelling at a speed of $v$ is given by,

  $KE=\frac{1}{2}m{v}^2$

And, the PE of an object of mass, m , due to its height, h , and gravity is given by,

  $PE=mgh$

Conservation of Mechanical Energy is given as,

  $K{E}_{before}+P{E}_{before}=K{E}_{after}+P{E}_{after}$

Where $KE$ is the kinetic energy and $PE$ is the potential energy.

Calculation:

Let the height of the rocket when its engine stops be $h$ and its additional height gained by $h{a}_{}$ . Then, using conservation of mechanical energy formula,

  $\begin{array}{c}K{E}_{before}+P{E}_{before}=K{E}_{after}+P{E}_{after}\\ 1960+mgh=0+mg\left(h+h_a\right)\\ 1960=mg{h}_a\end{array}$

Using the given values in above,

  $\begin{array}{c}1960=\left(10\right)\left(9.8\right)h_a\\ h_a=20\text{m}\end{array}$

Conclusion:

Thus, the additional height gained by the rocket is $20\text{m}$ .