If a projectile is shot at an angle of pi/3 and at a velocity of 17 m/s, when will it reach its maximum height??

Dec 4, 2016

The answer is $= 11.1 m$

Explanation:

At the maximum height, the vertical component of the velocity $= 0$

So, we use the equation

${v}^{2} = {u}^{2} + 2 a s$

$v = 0$

$u = {u}_{0} \sin \theta$

$a = - g = - 9.8 m {s}^{- 2}$

$s = h$ is the maximum height

$0 = {\left({u}_{0} \sin \theta\right)}^{2} - 2 g h$

$h = {\left({u}_{0} \sin \theta\right)}^{2} / \left(2 g\right)$

$= {\left(17 \cdot \sin \left(\frac{\pi}{3}\right)\right)}^{2} / \left(2 \cdot 9.8\right)$

$\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

$h = \frac{{17}^{2} \cdot \frac{3}{4}}{19.6} = 11.1 m$