# If a projectile is shot at an angle of (5pi)/12 and at a velocity of 19 m/s, when will it reach its maximum height??

Apr 1, 2017

The projectile will reach its maximum height when the velocity equals zero.

#### Explanation:

At an angle of $\left(\frac{5 \pi}{12}\right)$, the initial velocity forms the hypotenuse of a right-angled triangle with angle $\left(\frac{5 \pi}{12}\right)$ at the base.
The maximum height of the shot will be the vertical component $\left(y\right)$, of the initial velocity vector of $19$ m/s, and will occur when the final velocity equals zero.
The initial velocity of the vertical component is $V o y = 19 \sin \left(\frac{5 \pi}{12}\right) = 0.434$ m/s.
Using $V = V o + a t$ we can rearrange to find $t$:

$t = \frac{V - V o}{a}$

$= \frac{0 - 0.434}{-} 9.8 = 0.04 s$

Assuming that the force due to gravity is in the downward direction, therefore, the upward direction has a negative sign.