# A projectile is shot from the ground at an angle of ( pi)/3  and a speed of 9 /4 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?

Dec 27, 2017

The distance is $= 0.3 m$

#### Explanation:

Resolving in the vertical direction ${\uparrow}^{+}$

The initial velocity is ${u}_{0} = \frac{9}{4} \sin \left(\frac{1}{3} \pi\right) m {s}^{-} 1$

Applying the equation of motion

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

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

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

Therefore,

The greatest height is ${h}_{y} = s = \frac{0 - {\left(\frac{9}{4} \sin \left(\frac{1}{3} \pi\right)\right)}^{2}}{- 2 g}$

${h}_{y} = {\left(\frac{9}{4} \sin \left(\frac{1}{3} \pi\right)\right)}^{2} / \left(2 g\right) = 0.194 m$

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

Applying the equation of motion

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

The time is $t = \frac{v - u}{- g} = \frac{0 - \frac{9}{4} \sin \left(\frac{1}{3} \pi\right)}{- 9.8} = 0.199 s$

Resolving in the horizontal direction ${\rightarrow}^{+}$

The velocity is constant and ${u}_{x} = \frac{9}{4} \cos \left(\frac{1}{3} \pi\right)$

The distance travelled in the horizontal direction is

${s}_{x} = {u}_{x} \cdot t = \frac{9}{4} \cos \left(\frac{1}{3} \pi\right) \cdot 0.199 = 0.224 m$

The distance from the starting point is

$d = \sqrt{{h}_{y}^{2} + {s}_{x}^{2}} = \sqrt{{0.194}^{2} + {0.224}^{2}} = 0.3 m$