# If a projectile is shot at an angle of (3pi)/8 and at a velocity of 2m/s, when will it reach its maximum height?

Jun 26, 2018

The time is $= 0.19 s$

#### Explanation:

The initial speed is $u = 2 m {s}^{-} 1$

The angle is $\theta = \frac{3}{8} \pi r a d$

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

The equation of the trajectory of the projectile is

$y = x \tan \theta - \frac{g {x}^{2}}{2 {u}^{2} {\cos}^{2} \theta}$..............$\left(1\right)$

At the maximum height

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\implies$, $\frac{\mathrm{dy}}{\mathrm{dx}} = \tan \theta - \frac{2 g x}{2 {u}^{2} \cos \theta}$

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

Therefore,

$\tan \theta - \frac{g x}{{u}^{2} {\cos}^{2} \theta} = 0$

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

The constant horizontal velocity is

${v}_{x} = u \cos \theta$

Therefore,

The time to reach the greatest height is

$t = \frac{x}{v} _ x = \left({u}^{2} \frac{\sin \theta \cos \theta}{g}\right) \cdot \frac{1}{u \cos \theta}$

$= \frac{u \sin \theta}{g}$

$= 2 \cdot \sin \frac{\frac{3}{8} \pi}{9.8}$

$= 0.19 s$