# If a projectile is shot at an angle of pi/2 and at a velocity of 37 m/s, when will it reach its maximum height??

Jan 5, 2017

The vertical component of the velocity at time $t$ is given by $v \left(t\right) = u - g t$, where $u$ is the projectile's initial vertical component of velocity and $g$ is the gravitational acceleration constant.
So we are required to solve $0 = u - g t$, which gives $t = \frac{u}{g}$
Now, $u = 37 \sin \left(\frac{\pi}{2}\right)$m/s and we take $g = 10 m b \otimes \left\{\right\}$m/s$\setminus m b \otimes {\left\{\right\}}^{2}$. So the final answer is $t = \frac{37}{10} = 3.7$s.