# If a projectile is shot at an angle of (5pi)/12 and at a velocity of 21 m/s, when will it reach its maximum height??

Jan 21, 2016

Given, equation for time taken by projectile to reach its maximum height is $t = \frac{{v}_{o} \sin \left(\setminus \theta\right)}{g}$

Given, ${v}_{o} = 21 m {s}^{- 1} \text{ and } \setminus \theta = \frac{5 \setminus \pi}{12}$ and we'll take $g = 9.8 m {s}^{- 2}$
We don't have an exact value of $\sin \left(\frac{5 \setminus \pi}{12}\right)$ but we can separate is such that we can solve it.
$\sin \left(\frac{5 \setminus \pi}{12}\right) = \sin \left(\frac{8 \setminus \pi}{12} - \frac{3 \setminus \pi}{12}\right) = \sin \left(\frac{2 \setminus \pi}{3} - \setminus \frac{\pi}{4}\right)$
Using the solution for $\sin$ functions $\sin \left(x + y\right) = \sin \left(x\right) \cos \left(y\right) + \cos \left(x\right) \sin \left(y\right)$ we get,
$\sin \left(\frac{2 \setminus \pi}{3} - \setminus \frac{\pi}{4}\right) = \sin \left(\frac{2 \setminus \pi}{3}\right) \cos \left(\frac{\pi}{4}\right) - \cos \left(\frac{2 \setminus \pi}{3}\right) \sin \left(\frac{\pi}{4}\right)$
$\text{i.e } \sin \left(\frac{2 \setminus \pi}{3} - \setminus \frac{\pi}{4}\right) = \frac{\setminus \sqrt{3} + 1}{2 \setminus \sqrt{2}}$

So, $t = 21 \cdot \frac{\setminus \sqrt{3} + 1}{2 \setminus \sqrt{2} \cdot 9.8} s$