# If a projectile is shot at an angle of (2pi)/3 and at a velocity of 5 m/s, when will it reach its maximum height?

Feb 22, 2018

time taken to reach maximum height = $0.4 s$ (approx)

#### Explanation:

The velocity vector of the projectile has a x- component and y - component .
since we are dealing with height , we just need the y-component .
velocity in $y$ direction = $5 \sin \left(\frac{2 \pi}{3}\right) m {s}^{-} 1$
=$\frac{5 \sqrt{3}}{2} m {s}^{-} 1$

now using initial velocity $u = \frac{5 \sqrt{3}}{2} m {s}^{-} 1$ ,

final velocity $v = 0 m {s}^{-} 1$, acceleration due to gravity $a = - 9.8 m {s}^{-} 2$ .

we have $v = u + a t$

$\implies 0 = \frac{5 \sqrt{3}}{2} + - 9.8 t$

$\implies 9.8 t = \frac{5 \sqrt{3}}{2}$

=> t = (5sqrt3) /(2×9.8)

$\implies t = 0.4 s$ (approx)