# If a projectile is shot at an angle of (7pi)/12 and at a velocity of 9 m/s, when will it reach its maximum height?

Apr 17, 2017

$0.63 \sec$

#### Explanation:

We know that according to the law of motion .
$V = U + a \cdot t$

$\therefore$ this law is true for the frame of reference for non inertial frame of reference .

So it can be written as ,
${V}_{y} = {U}_{y} + {a}_{y} \cdot t$

we know that vertical component of the velocity is
$U \sin \theta$
here it can be written as

$U \sin \left(\frac{7 \pi}{12}\right) = U \sin \left(\pi - \frac{3 \pi}{12}\right) = U \sin \left(\frac{\pi}{4}\right)$as sin is positive in 2 quadrant

for attaining maximum height we get final velocity ${V}_{y} = 0$ so upon substituting the values in the equation we get .
$0 = 9 \cdot \frac{1}{\sqrt{2}} - 10 t$

as the acceleration due to gravity is in negative direction.
so we get .

$\frac{9}{10 \cdot \sqrt{2}} = t$
$t = \frac{9 \sqrt{2}}{20} = \frac{12.69}{2} = 0.63 \sec$