# A projectile is shot at an angle of (5pi)/12  and a velocity of  17 m/s. How far away will the projectile land?

May 3, 2016

$14.45 m$

#### Explanation:

Let us consider that the $\text{Velocity of projection"=u=17m/s and "Angle of projection} = \alpha = 5 \frac{\pi}{12}$

The vertical component of the velocity =$u \sin \alpha$

The horizontal component of the velocity =$u \cos \alpha$

Let us also consider that the time of flight of the projectile is T

This T time will be required to return the prrojectile to the ground i.e.after this "T"time net vertical displacement is zero

So we can write

$0 = u \sin \alpha \cdot T - \frac{1}{2} \cdot g \cdot {T}^{2}$
$T = \frac{2 u \sin \alpha}{g}$

Horizontal displacement = distance from point of projection to the point of landing $= \text{Horizotal component of velocity} \times T = u \cos \alpha \times \frac{2 u \sin \alpha}{g} = \frac{{u}^{2} \sin 2 \alpha}{g} = \frac{{17}^{2} \sin \left(2 \cdot 5 \frac{\pi}{12}\right)}{10} m = \left(28.9 \cdot \sin \left(\pi - \frac{\pi}{6}\right)\right) = 28.9 \cdot \sin \left(\frac{\pi}{6}\right) = \frac{28.9}{2} = 14.45 m$