If a projectile is shot at an angle of #pi/3# and at a velocity of #32 m/s#, when will it reach its maximum height??

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Answer 1 · 2016-05-24T07:25:51

at time #t=2.827838053" "#seconds

The equation for the vertical displacement for projectiles

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

The equation for the horizontal displacement for projectiles is

#x=v_0 cos theta *t#
Maximizing the vertical displacement by obtaining the first derivative of y with respect to time t

#dy/dt=v_0 sin theta+1/2*g*2*t#

set #dy/dt=0#

#v_0 sin theta+1/2*g*2*t=0#

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

#t=(-32 sin ((pi)/3))/(-9.8)#

#t=(80sqrt3)/49#

#t=2.827838053" "#seconds

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