# If a projectile is shot at an angle of (3pi)/8 and at a velocity of 12 m/s, when will it reach its maximum height?

Jul 23, 2017

$t = 1.1 \sec s$

#### Explanation:

Maximum height is reached when the vertical component of velocity $= 0$

the initial vertical component " u=12sin((3pi)/8)

we will use

$v = u + g t \uparrow$

$= 12 \sin \left(\frac{3 \pi}{8}\right) - 10 t$

$t = \frac{12}{10} \sin \left(\frac{3 \pi}{8}\right)$

$t = 1.1 \sec s$