Chapter 8, Problem 7P

To determine

To sketch: The graph of the projectile with initial speed ${\nu }_0=32\text{ft/s}$, fired into a headwind of $w=24\text{ft/s}$ for different angles.

Explanation of Solution

The initial speed of the projectile is ${\nu }_0=32\text{ft/s}$, fired into a headwind of $w=24\text{ft/s}$ for $\theta =5°,15°,30°,40°,45°,55°,60°$ and $75°$.

The parametric equations for the path of a particle with gravity with an angle $\theta$ are,

$x=\left({\nu }_0\cos \theta \right)t$

And

$y=\left({\nu }_0\sin \theta \right)t-\frac{1}{2}g{t}^2$

The headwind w slows down the horizontal component velocity only because headwind only affects the horizontal component.

$\begin{array}{c}x=\left({\nu }_0\cos \theta \right)t-wt\\ y=\left({\nu }_0\sin \theta \right)t-\frac{1}{2}g{t}^2\end{array}$

Substitute 32 ft/s for ${\nu }_0$, 24 ft/s for $w$,32 ft/s for g in the above expressions.

$x=\left(32\cos \theta \right)t-24t$ (1)

And

$\begin{array}{c}y=\left(32\sin \theta \right)t-\frac{1}{2}\times 32t^2\\ y=\left(32\sin \theta \right)t-16t^2\end{array}$ (2)

Calculate the value of t from equation (1).

$\begin{array}{c}x=\left(32\cos \theta \right)t-24t\\ =t\left(32\cos \theta -24\right)\\ t=\frac{x}{32\cos \theta -24}\end{array}$

Substitute $\frac{x}{32\cos \theta -24}$ for t in equation (2).

$y=\left(32\sin \theta \right)\left(\frac{x}{32\cos \theta -24}\right)-16{\left(\frac{x}{32\cos \theta -24}\right)}^2$

Plot the graph by the above equation with different values of $\theta$.

Figure (1)

Thus, Figure (1) shows that the maximum horizontal distance is at $23°$.