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

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

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

If a projectile is shot at an angle of (5pi)/12(5pi)/12 and at a velocity of 27 m/s27 m/s, when will it reach its maximum height?