If a projectile is shot at a velocity of 15 m/s and an angle of pi/6, how far will the projectile travel before landing?

May 16, 2018

The horizontal distance is $= 19.88 m$

Explanation:

The trajectory of a projectile in the $x - y$ plane is given by the equation

$y = x \tan \theta - \frac{g {x}^{2}}{2 {u}^{2} {\cos}^{2} \theta}$

The initial velocity is $u = 15 m {s}^{-} 1$

The angle is $\theta = \frac{\pi}{6}$

The acceleration due to gravity is $g = 9.8 m {s}^{-} 2$

The vertical distance travelled is $y = 0$

Therefore,

$x \tan \left(\frac{\pi}{6}\right) - \frac{9.8 {x}^{2}}{2 \cdot {15}^{2} {\cos}^{2} \left(\frac{\pi}{6}\right)} = 0$

$0.577 x - 0.029 {x}^{2} = 0$

$x \left(0.577 - 0.029 x\right) = 0$

$x = 0$ which is the initial point

$x = \frac{0.577}{0.029} = 19.88 m$

graph{0.577x-0.029x^2 [-0.96, 21.54, -4.6, 6.65]}