If a projectile is shot at an angle of #(pi)/8# and at a velocity of #21 m/s#, when will it reach its maximum height?

1 Answer
Apr 1, 2018

The time is #=0.82s#

Explanation:

Resolving in the vertical direction #uarr^+#

The initial velocity is #u_0=(21)sin(1/8pi)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-(21)sin(1/8pi))/(-9.8)=0.82s#