If a projectile is shot at an angle of $\frac{π}{3}$ $pi/3$ and at a velocity of $17 \frac{m}{s}$ $17 m/s$ , when will it reach its maximum height??

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Answer 1 · 2016-12-04T07:25:34

The answer is $= 11.1 m$ $=11.1m$

At the maximum height, the vertical component of the velocity $= 0$ $=0$

So, we use the equation

$v^{2} = u^{2} + 2 a s$ $v^2=u^2+2as$

$v = 0$ $v=0$

$u = u_{0} sin θ$ $u=u_0sintheta$

$a = - g = - 9.8 m s^{- 2}$ $a=-g=-9.8ms^(-2)$

$s = h$ $s=h$ is the maximum height

$0 = {(u_{0} sin θ)}_{0}^{2} - 2 g h$ $0=(u_0sintheta)^2-2gh$

$h = \frac{{(u_{0} sin θ)}_{0}^{2}}{2 g}$ $h=(u_0sintheta)^2/(2g)$

$= \frac{{(17 \cdot sin (\frac{π}{3}))}^{2}}{2 \cdot 9.8}$ $=(17*sin(pi/3))^2/(2*9.8)$

$sin (\frac{π}{3}) = \frac{\sqrt{3}}{2}$ $sin(pi/3)=sqrt3/2$

$h = \frac{17^{2} \cdot \frac{3}{4}}{19.6} = 11.1 m$ $h=(17^2*3/4)/19.6=11.1m$

If a projectile is shot at an angle of π3pi/3 and at a velocity of 17ms17 m/s, when will it reach its maximum height??