# If a projectile is shot at a velocity of 6 m/s and an angle of pi/4, how far will the projectile travel before landing?

Jul 30, 2016

$3.67 m$

#### Explanation:

Let the velocity of projection of the projectile 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
$h = u \sin \alpha \times T + \frac{1}{2} g {T}^{2}$
$\implies 0 = u \times T - \frac{1}{2} \times g \times {T}^{2}$
$\text{where " g= "acceleration ""due " "to " } g r a v i t y$

$\therefore T = \frac{2 u \sin \alpha}{g}$

The horizontal displacement during this T sec is $R = u \cos \alpha \times T = \frac{{u}^{2} \sin \left(2 \alpha\right)}{g}$

In the given problem

$u = 6 \frac{m}{s} , g = 9.8 \frac{m}{s} ^ 2 \mathmr{and} \alpha = \frac{\pi}{4}$

So the distance travaresed by the projectile before landing is

$R = \frac{{6}^{2} \cdot \sin \left(2 \cdot \frac{\pi}{4}\right)}{9.8} m \approx 3.67 m$