If a projectile is shot at an angle of #pi/3# and at a velocity of #4 m/s#, when will it reach its maximum height??

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Answer 1 · 2018-03-20T09:29:00

The time is #=0.88s#

Resolving in the vertical direction #uarr^+#

The initial velocity is #u_0=(4)sin(1/3pi)ms^-1#

At the greatest height, #v=0ms^-1#

The acceleration due to gravity is #a=-g=-9.8ms^-2#

The time to reach the greatest height is #=ts#

Applying the equation of motion

#v=u+at=u- g t #

The time is #t=(v-u)/(-g)=(0-(10)sin(1/3pi))/(-9.8)=0.88s#