Question

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??

Answer 1

Given, equation for time taken by projectile to reach its maximum height is `t=(v_osin(\theta))/g`

Given, `v_o=21ms^{-1}" and " \theta=(5\pi)/12` and we'll take `g=9.8ms^{-2}`
We don't have an exact value of `sin((5\pi)/12)` but we can separate is such that we can solve it.
`sin((5\pi)/12)=sin((8\pi)/12-(3\pi)/12)=sin((2\pi)/3-\pi/4)`
Using the solution for `sin` functions `sin(x+y)=sin(x)cos(y)+cos(x)sin(y)` we get,
`sin((2\pi)/3-\pi/4)=sin((2\pi)/3)cos(pi/4)-cos((2\pi)/3)sin(pi/4)`
`"i.e "sin((2\pi)/3-\pi/4)=(\sqrt{3}+1)/(2\sqrt{2})`

So, `t=21*(\sqrt{3}+1)/(2\sqrt{2}*9.8)s`

Finalizing the answer is left as an exercise to the reader.