If a projectile is shot at an angle of #(5pi)/12# and at a velocity of #27 m/s#, when will it reach its maximum height?

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Answer 1 · 2017-07-27T05:21:47

The time is #=2.66s#

Resolving in the direction #uarr^+#

We apply the equation of motion

#v=u+at#

The initial velocity is #u=27sin(5/12pi)ms^-1#

The final velocity (at the maximum height) is #v=0ms^-1#

The acceleration is #a=-gms^-2#

Therefore,

the time is

#t=(v-u)/(g)#

#t=0-27sin(5/12pi)/(-9.8)=2.66s#