8.1.1 Projectile Motion Revisited

We found in Chapter 6 that the gravitational force on an object near the surface of a planet is $\vec{F}_{G} = (m g, down)$ . For a coordinate system with a vertical y-axis,

\vec{F}_{G} = - m g ĵ

(8.1)

Consequently, from Newton’s second law, the acceleration is

\begin{array}{l} a_{x} = \frac{{(F_{G})}_{x}}{m} = 0 \\ a_{y} = \frac{{(F_{G})}_{y}}{m} = - g \end{array}

(8.2)

Equations 8.2 justify the analysis of projectile motion in 《 SECTION 4.2—a downward acceleration $a_{y} = - g$ with no horizontal acceleration—where we found that a drag-free projectile follows a parabolic trajectory. The fact that the two components of acceleration are independent of each other allows us to solve for the vertical and horizontal motions.

However, the situation is quite different for a low-mass projectile, where the effects of drag are too large to ignore. We’ll leave it as a homework problem for you to show that the acceleration of a projectile subject to drag, with drag coefficient $C_{d}$ , is

\begin{array}{l} a_{x} = - \frac{ρ C_{d} A}{2 m} v_{x} \sqrt{v_{x}^{2} + v_{y}^{2}} \\ a_{y} = - g - \frac{ρ C_{d} A}{2 m} v_{y} \sqrt{v_{x}^{2} + v_{y}^{2}} \end{array}

(8.3)

Here the components of acceleration are neither constant nor independent of each other because $a_{x}$ depends on $v_{y}$ and vice versa. It turns out that these two equations cannot be solved exactly for the trajectory, but they can be solved numerically. FIGURE 8.2 shows the numerical solution for the motion of a 5 g plastic ball that’s been hit with an initial speed of $25 m /s$ . It doesn’t travel very far (the maximum distance without drag would be more than 60 m), and the maximum range is no longer reached for a launch angle of $45 °$ . Notice that the trajectories are not at all parabolic.

The figure illustrates the curved trajectories of a projectile affected by drag launched at three different angles graphed on the x y coordinate plane.

FIGURE 8.2 Full Alternative Text