Projectile Range

physicsmechanics/kinematicsdifficulty 3/5#projectile#kinematics#parabola

A projectile is launched from the ground with speed $v_0 = 15\ \text{m/s}$ at an angle
$\theta = 30^\circ$ above the horizontal. Taking $g = 9.8\ \text{m/s}^2$, its trajectory is

$$ y(x) = x\tan\theta - \frac{g\,x^2}{2\,v_0^2\cos^2\theta}. $$

Estimate the horizontal range $R$ — the value of $x$ at which the projectile returns to the
ground ($y = 0$).

5 10 15 20 -0.5 0.5 1 1.5 2 2.5 3 x (m) y (m)
Trajectory $y(x)$, in metres.

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Answer

$ 19.9 m $

Solution

The range of a projectile launched from the ground is
$R = \dfrac{v_0^2 \sin 2\theta}{g}$. With $v_0 = 15\ \text{m/s}$, $\theta = 30^\circ$ and
$g = 9.8\ \text{m/s}^2$,
$$R = \frac{15^2 \sin 60^\circ}{9.8} \approx 19.9\ \text{m}.$$