Chapter 8.7, Problem 16E

To determine

To find: the relationship for the given situation.

Answer to Problem 16E

The time of stay of projectile is directly proportional to the value of $\theta$ .

The horizontal distance increases as $\theta$ increases from $0°$ to $45°$ , and then horizontal distance decreases as $\theta$ increases from $45°$ to $90°$ .

Explanation of Solution

Given information:

The angle at which a projectile is launched, the time its stays in the air and the distance it covers.

Calculation:

When the projectile is in air, the vertical displacement ( y ) becomes zero. The time of projectile stay in the air is solved as follows:

  $\begin{array}{c}y=t\left|\overrightarrow{v}\right|\sin \theta -\frac{1}{2}g{t}^2\\ 0=t\left(\left|\overrightarrow{v}\right|\sin \theta -\frac{1}{2}gt\right)\left[\text{Put }y=0\right]\\ \left|\overrightarrow{v}\right|\sin \theta -\frac{1}{2}gt=0\\ \left|\overrightarrow{v}\right|\sin \theta =\frac{1}{2}gt\\ t=\frac{2\left|\overrightarrow{v}\right|\sin \theta }{g}\end{array}$

Thus time of stay of projectile is directly proportional to the value of $\theta$ ; i.e. greater the angle greater the projectile stays in the air.

The horizontal distance ( x ) for the given situation can be evaluated as follows:

  $\begin{array}{c}x=t\left|\overrightarrow{v}\right|\cos \theta \\ =\left(\frac{2\left|\overrightarrow{v}\right|\sin \theta }{g}\right)\left(\left|\overrightarrow{v}\right|\cos \theta \right)\left[\text{Put the value of }t\right]\\ =\frac{{\left|\overrightarrow{v}\right|}^2\sin 2\theta }{g}\end{array}$

The horizontal distance increases as $\theta$ increases from $0°$ to $45°$ , and then horizontal distance decreases as $\theta$ increases from $45°$ to $90°$ .