# A projectile is shot from the ground at a velocity of 12 m/s at an angle of pi/6. How long will it take for the projectile to land?

May 16, 2016

1.2s

#### 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 canh 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 = "acceleration" "due to" gravity”
$\therefore T = \frac{2 u \sin \alpha}{g}$
In our problem $u = 12 \frac{m}{s} \mathmr{and} \alpha = \frac{\pi}{6}$

Hence $T = \frac{2 \cdot 12 \sin \left(\frac{\pi}{6}\right)}{9.8}$
$= \frac{2 \times 12 \times \frac{1}{2}}{9.8} = 1.2 s$