Suppose a heavy ball is launched exactly horizontally at height h above a horizontal field. At the exact instant that the ball is launched, a second ball is simply dropped from height h. Which ball hits the ground first?

It may seem hard to believe, but—if air resistance is neglected—the balls hit the ground simultaneously. They do so because the horizontal and vertical components of projectile motion are independent of each other. The initial horizontal velocity of the first ball has no influence over its vertical motion. Neither ball has any initial motion in the vertical direction, so both fall distance h in the same amount of time. You can see this in FIGURE 4.13.

FIGURE 4.13 A projectile launched horizontally falls in the same time as a projectile that is released from rest.

FIGURE 4.14a shows a useful way to think about the trajectory of a projectile. Without gravity, a projectile would follow a straight line. Because of gravity, the particle at time t has “fallen” a distance ${\displaystyle \frac{1}{2}}g{t}^2$ below this line. The separation grows as ${\displaystyle \frac{1}{2}}g{t}^2$, giving the trajectory its parabolic shape.

FIGURE 4.14 A projectile follows a parabolic trajectory because it “falls” a distance $\frac{1}{2}g{t}^2$ below a straight-line trajectory.

Use this idea to think about the following “classic” problem in physics:

A hungry bow-and-arrow hunter in the jungle wants to shoot down a coconut that is hanging from the branch of a tree. He points his arrow directly at the coconut, but as luck would have it, the coconut falls from the branch at the exact instant the hunter releases the string. Does the arrow hit the coconut?

You might think that the arrow will miss the falling coconut, but it doesn’t. Although the arrow travels very fast, it follows a slightly curved parabolic trajectory, not a straight line. Had the coconut stayed on the tree, the arrow would have curved under its target as gravity caused it to fall a distance $\frac{1}{2}g{t}^2$ below the straight line. But $\frac{1}{2}g{t}^2$ is also the distance the coconut falls while the arrow is in flight. Thus, as FIGURE 4.14b shows, the arrow and the coconut fall the same distance and meet at the same point!