Chapter 13, Problem 21PT

To determine

To find: the angle is needed in order for the rocket to reach a range of 25 meters.

Answer to Problem 21PT

  $\theta \approx {18.89}^{\circ }.$

Explanation of Solution

Given information: A model rocket is launched with an initial velocity of 20 meters per

second. The range of a projectile is given by the formula $R=\frac{v^2}{g}\sin 2\theta$ where R is the range, v is the initial velocity, g is acceleration due to gravity or 9.8 meters per second squared and $\theta$ is the launch angle.

Calculation:

Given formula:

  $\begin{array}{c}R=\frac{v^2}{g}\sin 2\theta \\ R=25\\ v=20\\ g=9.8\\ 25=\frac{{20}^2}{9.8}\sin 2\theta \\ \sin 2\theta =\frac{9.8\times 25}{20\times 20}=\frac{245}{400}=0.6125\\ 2\theta ={\sin }^{-1}(0.6125)\approx 37.77\\ \theta \approx \frac{37.77}{2}\approx 18.89\\ \theta \approx {18.89}^{\circ }.\end{array}$