If a projectile is shot at a velocity of 9 m/s and an angle of pi/6, how far will the projectile travel before landing?

May 11, 2016

7.1m

Explanation:

Let the velocity of projection of the object be u with angle of projection $\alpha$ with the horizontal direction.
The vertical component of the velocity of projection is $u \sin \alpha$ and the horizontal component is $u \cos \alpha$

Now if the time of flight be T then the object will return to the ground after T sec and during this T sec its total vertical dosplacement h will be zero. So applying the equation of motion under gravity we can write
$h = u \sin \alpha \times T + \frac{1}{2} g {T}^{2} \implies 0 = u \times T - \frac{1}{2} \times g \times {T}^{2}$ where $g = \text{acceleration due to gravity}$
$\therefore T = \frac{2 u \sin \alpha}{g}$
The horizontal displacement during this T sec is $R = u \cos \alpha \times T = \frac{{u}^{2} \sin \left(2 \alpha\right)}{g}$

In our problem u=9m/s;alpha=pi/6 "and" g=9.8m/s^2

So the distance travelled by the projectile before landing is

$\text{Distance} = \frac{{9}^{2} \sin \left(2 \cdot \frac{\pi}{6}\right)}{9.8} = 7.1 m$