Problems Plus 1: Projectile Fired from the Origin

Full written solution to parts (a), (b) and (c) — mirrored from Vector Functions: Problems Plus, which also covers Problems 2–7.

A projectile is fired from the origin with angle of elevation α and initial speed v₀.

Assuming that air resistance is negligible and that the only force acting on the projectile is gravity, g, we showed in my earlier video that the position vector of the projectile is:

We also showed that the maximum horizontal distance (or range) of the projectile is achieved when α = 45° and in this case the range is R = (v₀)²/ g.

(a) At what angle should the projectile be fired to achieve maximum height and what is the maximum height?

(b) Fix the initial speed v₀ and consider the parabola x² + 2Ry - R² = 0, whose graph is shown in the figure below.

Show that the projectile can hit any target inside or on the boundary of the region bounded by the parabola and the x-axis, and that it can't hit any target outside this region.

(c) Suppose that the gun is elevated at an angle of inclination α in order to aim at a target that is suspended at a height h directly over a point D units downrange.

The target is released at the instant the gun is fired.

Show that the projectile always hits the target, regardless of the value v₀, provided the projectile does not hit the ground "before" D.

Solution to (a)

The projectile reaches maximum height when:

This is the maximum height attained when the projectile is fired with an angle of elevation α.

This maximum height is largest when α = 90°, which makes sense as it means the launch angle is straight up vertically!

In that case, sin α = 1 and the maximum height is (v₀)²/(2g).

Solution to (b)

We are asked to consider the parabola:

We can rearrange the parabola as an equation for y.

The points on or inside this parabola are those for which:

Recall my earlier videos on the formula for the position of a projectile:

When the projectile is fired at an angle of elevation α, the points (x, y) along its path satisfy the relations:

Recall the sine double angle identity from my earlier video.

For t in the specified range, we also have:

Now if we subtract this formula for the projectile trajectory by the formula for the parabola, we get:

Thus we have shown that every target that can be hit by the projectile lies on or inside the parabola.

Recall the Pythagorean Identity and the corresponding tangent and secant identity from my earlier videos.

Now let (a, b) be any point on or inside the parabola:

We seek an angle α such that (a, b) lies in the path of the projectile; that is, we wish to find an angle α such that:

Rearranging this equation we get:

This quadratic equation for tan α has real solutions exactly when the discriminant is nonnegative.

Recall my earlier video on the Quadratic Formula.

This condition is satisfied since (a, b) is on or inside the parabola:

It follows that (a, b) lies in the path of the projectile when tan α satisfies (*), that is, when:

Solution to (c)

If the gun is pointed at a target with height h at a distance D downrange, then:

When the projectile reaches a distance D downrange (remember we are assuming that it doesn't hit the ground first), we have:

Meanwhile, the target, whose x-coordinate is also D, has fallen from height h to height:

Thus the projectile hits the target!