The path of a projectile

Eliminating t to find the Cartesian equation of a projectile's path, the general range formula and the angle of maximum range, finding the launch angle needed to hit a target, and projectiles launched from a height.

The equation of the path

From $x = Vt\cos\theta$, substitute $t = \dfrac{x}{V\cos\theta}$ into $y = Vt\sin\theta - \tfrac12 gt^2$:

Path of a projectile

$$y = x\tan\theta - \frac{gx^2}{2V^2\cos^2\theta} = x\tan\theta - \frac{gx^2}{2V^2}\big(1 + \tan^2\theta\big)$$

The path is a concave-down parabola. The second form uses $\sec^2\theta = 1 + \tan^2\theta$, which turns the equation into a quadratic in $\tan\theta$.

Launch a projectile

Speed V in m/s and angle a in degrees, with g = 10. Try angles that add to 90°, such as 30° and 60°: they give the same range.

Try this

For a fixed speed, which angle gives the greatest range?

Range

Range on level ground

$$R = \frac{V^2\sin 2\theta}{g}$$

Since $\sin 2\theta \le 1$, the greatest range is $\dfrac{V^2}{g}$, at $\theta = 45^\circ$. Complementary angles $\theta$ and $90^\circ - \theta$ give the same range.

Example 1●

Find the greatest range of a ball thrown at $20$ m/s, with $g = 10$.

$R_{\max} = \dfrac{20^2}{10} = 40$ m, at $45^\circ$.

Hitting a target

Example 2●●●

A ball is thrown at $20$ m/s, with $g = 10$. At what angles must it be thrown to pass through the point $(20, 5)$?

Let $T = \tan\theta$. Then $5 = 20T - \dfrac{10 \times 400}{800}(1 + T^2) = 20T - 5 - 5T^2$.

So $T^2 - 4T + 2 = 0$, giving $T = 2 \pm \sqrt2$.

$\theta \approx 73.7^\circ$ or $\theta \approx 30.4^\circ$.

Flashcards

Path of a projectile

$y = x\tan\theta - \dfrac{gx^2}{2V^2}(1 + \tan^2\theta)$

Range on level ground

$\dfrac{V^2\sin 2\theta}{g}$

Angle for maximum range

$45^\circ$

Practice

Find the greatest range, in metres, of a ball thrown at $20$ m/s, with $g = 10$.
Find the range at $60^\circ$ for the same speed, correct to two decimal places.
A ball thrown at $30^\circ$ has the same range as a ball thrown at the same speed at which other angle, in degrees?
Solve $T^2 - 4T + 2 = 0$. Enter both answers separated by a comma.
A projectile has path $y = x - \dfrac{x^2}{40}$. Find $y$ when $x = 10$.
Find the range of the path $y = x - \dfrac{x^2}{40}$.
The path of a projectile is:
To find the launch angle that hits a point, the path equation is solved as a quadratic in:
The quadratic for the launch angle has no real solutions. This means: