# A projectile is shot from the ground at a velocity of 3 m/s at an angle of pi/3. How long will it take for the projectile to land?

Apr 25, 2018

The time is $= 1.06 s$

#### Explanation:

The equation describing the trajectory of the projectile in the $x - y$ plane is

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

The initial velocity is $u = 3 m {s}^{-} 1$

The angle is $\theta = \left(\frac{1}{3} \pi\right) r a d$

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

The distance $y = 0$

Therefore,

$x \tan \left(\frac{1}{3} \pi\right) - \frac{9.8 \cdot {x}^{2}}{2 \cdot {3}^{2} {\cos}^{2} \left(\frac{\pi}{3}\right)} = 0$

$1.732 x - 1.089 {x}^{2} = 0$

$x \left(1.732 - 1.089 x\right) = 0$

$x = 0$, this is the starting point

$x = \frac{1.732}{1.089} = 1.59 m$

The time is $t = \frac{x}{v} _ x = \frac{1.59}{3 \cos \left(\frac{\pi}{3}\right)} = 1.06 s$