# A projectile is shot from the ground at an angle of pi/8  and a speed of 9 m/s. When the projectile is at its maximum height, what will its distance from the starting point be?

Dec 5, 2017

The distance is $= 2.97 m$

#### Explanation:

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

The initial velocity is ${u}_{0} = 9 \sin \left(\frac{1}{8} \pi\right) m {s}^{-} 1$

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 ${h}_{y} = s = \frac{0 - {\left(9 \sin \left(\frac{1}{8} \pi\right)\right)}^{2}}{- 2 g}$

${h}_{y} = {\left(9 \sin \left(\frac{1}{8} \pi\right)\right)}^{2} / \left(2 g\right) = 0.61 m$

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

Applying the equation of motion

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

The time is $t = \frac{v - u}{- g} = \left(0 - 9 \sin \left(\frac{1}{8} \pi\right)\right) \left(- 9.8\right) = 0.35 s$

Resolving in the horizontal direction $\rightarrow$

The velocity is constant and ${u}_{x} = 9 \cos \left(\frac{1}{8} \pi\right)$

The distance travelled in the horizontal direction is

${s}_{x} = {u}_{x} \cdot t = 9 \cos \left(\frac{1}{8} \pi\right) \cdot 0.35 = 2.91 m$

The distance from the starting point is

$d = \sqrt{{h}_{y}^{2} + {s}_{x}^{2}} = \sqrt{{0.61}^{2} + {2.91}^{2}} = 2.97 m$