Chapter 6.2, Problem 122AYU

For Problems 121-124, use the following discussion.

Projectile Motion The path of a projectile fired at an inclination $\theta$ to the horizontal with initial speed $v_0$ is a parabola (see the figure).

The range $R$ of the projectile, that is, the horizontal distance that the projectile travels, is found by using the function

$R\left(\theta \right)=\frac{v_0^2\sin \left(2\theta \right)}{g}$

where $g\approx 32.2$ feet per second per second $\approx 9.8$ meters per second per second is the acceleration due to gravity. The maximum height $H$ of the projectile is given by the function

$H\left(\theta \right)={\frac{v_0^2\left(\sin \theta \right)}{2g}}^2$

In Problems 121-124, find the range $R$ and maximum height $H$. (See the discussion on the previous page.)

The projectile is fired at an angle of ${30}^{\circ }$ to the horizontal with an initial speed of 150 meters per second.

To determine

To find: The range $\text{R}$ and maximum height $\text{H}$.

Answer to Problem 122AYU

Solution:

$R\left(\theta \right)=1988.32$ meter, $H\left(\theta \right)=286.99$ meter

Explanation of Solution

Given:

$\theta =30°,v_0=150$ meters per second, $g\approx 9.8$ meter per second

Calculation:

$R\left(\theta \right)=\frac{v_0^2\sin 2\theta }{g}=\frac{{150}^2\sin 2\left(30°\right)}{9.8}=1988.32$ meter

$H\left(\theta \right)=\frac{v_0^2{\left(\sin \theta \right)}^2}{2g}=\frac{{150}^2{\left(\sin \left(30°\right)\right)}^2}{2\left(9.8\right)}=286.99$ meter