# If a projectile is shot at an angle of (3pi)/8 and at a velocity of 7 m/s, when will it reach its maximum height?

Mar 13, 2018

The projectile will reach the maximum height at time $t = \left(\frac{7}{10}\right) . \cos \left(\frac{\pi}{8}\right)$ seconds after being thrown.

#### Explanation:

The initial velocity of the projectile is inclined at an angle $\theta$ with the horizontal.
Therefore the vertical motion which can be treated independently from the horizontal motion...will reach a point where its final velocity is zero.

and that instant is the position of maximum height.

The initial velocity component in the vertical direction is $u \cdot \cos \left(\frac{\pi}{8}\right)$.

So, we have

$v = u \cdot \cos \left(\frac{\pi}{8}\right) - g \cdot t$

where $v$ is the final velocity, which is zero.

Therefore, the projectile will reach the maximum height at a

$t = \frac{u \cdot \cos \left(\frac{\pi}{8}\right)}{g}$

Substituting the values of $u$ and $g$, we have

$t = \left(\frac{7}{10.}\right) \cos \left(\frac{\pi}{8}\right)$