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

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Answer 1 · 2017-07-27T05:21:47

The time is $= 2.66 s$ $=2.66s$

Resolving in the direction ${↑ ⏐ ⏐}^{+}$ $uarr^+$

We apply the equation of motion

$v = u + a t$ $v=u+at$

The initial velocity is $u = 27 sin (\frac{5}{12} π) m s^{- 1}$ $u=27sin(5/12pi)ms^-1$

The final velocity (at the maximum height) is $v = 0 m s^{- 1}$ $v=0ms^-1$

The acceleration is $a = - g m s^{- 2}$ $a=-gms^-2$

Therefore,

the time is

$t = \frac{v - u}{g}$ $t=(v-u)/(g)$

$t = 0 - 27 \frac{sin (\frac{5}{12} π)}{- 9.8} = 2.66 s$ $t=0-27sin(5/12pi)/(-9.8)=2.66s$

If a projectile is shot at an angle of 5π12(5pi)/12 and at a velocity of 27ms27 m/s, when will it reach its maximum height?