# A projectile is shot from the ground at a velocity of 8 m/s and at an angle of (2pi)/3. How long will it take for the projectile to land?

Jun 27, 2018

$| \vec{v} | = 8 m \cdot {s}^{- 1}$

${\vec{v}}_{x} = 8 \cos \left(\frac{2 \pi}{3}\right) m \cdot {s}^{- 1} \hat{i} = - 4 \hat{i}$

${\vec{v}}_{y} = 8 \sin \left(\frac{2 \pi}{3}\right) m \cdot {s}^{- 1} \hat{j} = 4 \sqrt{3} \hat{j}$

In y axis,

${v}_{i} = 4 \sqrt{3}$

At maximum height,

${v}_{f} = 0$

$a = - 10 m \cdot {s}^{- 2}$

Applying first equation of motion,

$0 = 4 \sqrt{3} - 10 t \Rightarrow t = 0.4 \cdot \sqrt{3} \Rightarrow t = 0.692 s$

So, time taken in reaching maximum height is 0.692 s.Time taken for projectile to fall from maximum height to ground is also 0.692s.

$\text{Total time taken} = 2 \cdot 0.692 = 1.384 s$