A projectile is shot from the ground at an angle of $\frac{5 π}{12}$ $(5 pi)/12$ and a speed of $5 \frac{m}{s}$ $5 m/s$ . Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

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Answer 1 · 2017-12-19T12:07:42

The distance is $= 2.24 m$ $=2.24m$

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

The initial velocity is $u_{0} = 5 sin (\frac{5}{12} π) m s^{- 1}$ $u_0=5sin(5/12pi)ms^-1$

Applying the equation of motion

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

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

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

Therefore,

The greatest height is $h_{y} = s = \frac{0 - {(5 sin (\frac{5}{12} π))}^{2}}{- 2 g}$ $h_y=s=(0-(5sin(5/12pi))^2)/(-2g)$

$h_{y} = \frac{{(5 sin (\frac{5}{12} π))}^{2}}{2 g} = 1.19 m$ $h_y=(5sin(5/12pi))^2/(2g)=1.19m$

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 $t = \frac{v - u}{- g} = \frac{0 - 5 sin (\frac{5}{12} π)}{- 9.8} = 1.47 s$ $t=(v-u)/(-g)=(0-5sin(5/12pi))/(-9.8)=1.47s$

Resolving in the horizontal direction $\to^{+}$ $rarr^+$

The velocity is constant and $u_{x} = 5 cos (\frac{5}{12} π)$ $u_x=5cos(5/12pi)$

The distance travelled in the horizontal direction is

$s_{x} = u_{x} \cdot t = 5 cos (\frac{5}{12} π) \cdot 1.47 = 1.90 m$ $s_x=u_x*t=5cos(5/12pi)*1.47=1.90m$

The distance from the starting point is

$d = \sqrt{h_{y}^{2} + s_{x}^{2}} = \sqrt{{1.19}^{2} + {1.90}^{2}} = 2.24 m$ $d=sqrt(h_y^2+s_x^2)=sqrt(1.19^2+1.90^2)=2.24m$

A projectile is shot from the ground at an angle of 5π12(5 pi)/12 and a speed of 5ms5 m/s. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?