If a projectile is shot at an angle of $\frac{π}{12}$ $pi/12$ and at a velocity of $2 \frac{m}{s}$ $2 m/s$ , when will it reach its maximum height??

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Answer 1 · 2017-07-19T16:38:51

The time to reach the greatest height is $= 0.053 s$ $=0.053s$

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

The initial velocity is $u_{y} = v sin θ = 2 \cdot sin (\frac{1}{12} π)$ $u_y=vsintheta=2*sin(1/12pi)$

The Acceleration is $a = - g$ $a=-g$

At the maximum height, $v = 0$ $v=0$

We apply the equation of motion

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

to calculate the time to reach the greatest height

$0 = 2 sin (\frac{1}{12} π) - g \cdot t$ $0=2sin(1/12pi)-g*t$

$t = \frac{2}{g} \cdot sin (\frac{1}{12} π)$ $t=2/(g)*sin(1/12pi)$

$= 0.053 s$ $=0.053s$

If a projectile is shot at an angle of pi/12π12pi/12 and at a velocity of 2 m/s2ms2 m/s, when will it reach its maximum height??