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??

1 Answer
Rohan A.K
Jan 21, 2016

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.

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