If a projectile is shot at an angle of $\frac{π}{8}$ $(pi)/8$ and at a velocity of $15 \frac{m}{s}$ $15 m/s$ , when will it reach its maximum height?

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Answer 1 · 2018-04-09T09:05:39

The time is $= 0.59 s$ $=0.59s$

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

The initial velocity is $u_{0} = (15) sin (\frac{1}{8} π) m s^{- 1}$ $u_0=(15)sin(1/8pi)ms^-1$

At the greatest height, $v = 0 m s^{- 1}$ $v=0ms^-1$

The acceleration due to gravity is $a = - g = - 9.8 m s^{- 2}$ $a=-g=-9.8ms^-2$

The time to reach the greatest height is $= t s$ $=ts$

Applying the equation of motion

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

The time is $t = \frac{v - u}{- g} = \frac{0 - (15) sin (\frac{1}{8} π)}{- 9.8} = 0.59 s$ $t=(v-u)/(-g)=(0-(15)sin(1/8pi))/(-9.8)=0.59s$

If a projectile is shot at an angle of (pi)/8π8(pi)/8 and at a velocity of 15 m/s15ms15 m/s, when will it reach its maximum height?