If a projectile is shot at an angle of #(7pi)/12# and at a velocity of #2 m/s#, when will it reach its maximum height?

1 Answer

Answer:

Time #t=(5sqrt6+5sqrt2)/98=0.1971277197" "#second

Explanation:

For the vertical displacement #y#

#y=v_0 sin theta * t+1/2*g*t^2#

We maximize displacement #y# with respect to #t#

#dy/dt=v_0 sin theta* dt/dt+1/2*g*2*t^(2-1)*dt/dt#

#dy/dt=v_0 sin theta+g*t#

set #dy/dt=0# then solve for #t#

#v_0 sin theta+g*t=0#

#t=(-v_0 sin theta)/g#

#t=(-2*sin ((7pi)/12))/(-9.8)#

Note: #sin ((7pi)/12)=sin ((5pi)/12)=(sqrt(6)+sqrt(2))/4#

#t=(-2*((sqrt(6)+sqrt(2)))/4)/(-9.8)#

#t=(5sqrt6+5sqrt2)/98=0.1971277197" "#second

God bless....I hope the explanation is useful.