# A projectile is shot from the ground at an angle of pi/8  and a speed of 2 m/s. When the projectile is at its maximum height, what will its distance from the starting point be?

Nov 23, 2016

${t}_{m} = \frac{2 \sin \left(\frac{\pi}{8}\right)}{g}$

$d = \frac{2 \sin \left(\frac{\pi}{8}\right) \left(4 {\cos}^{2} \left(\frac{\pi}{8}\right) + {\sin}^{2} \left(\frac{\pi}{8}\right)\right)}{g}$

#### Explanation:

${a}_{y} \left(t\right) = - g$.

${v}_{y} \left(t\right) = \int {a}_{y} \left(t\right) \mathrm{dt} = - g t + {v}_{0 y} = - g t + {v}_{0} \sin \left(\theta\right)$.

$y \left(t\right) = \int {v}_{y} \left(t\right) = - \frac{1}{2} g {t}^{2} + {v}_{0} \sin \left(\theta\right) t$.

${a}_{x} \left(t\right) = 0$.

${v}_{x} \left(t\right) = \int {a}_{x} \left(t\right) = 0 + {v}_{0 x} = {v}_{0} \cos \left(\theta\right)$.

$x \left(t\right) = \int {v}_{x} \left(t\right) = {v}_{0} \cos \left(\theta\right) t$.

The projectile is at its maximum height when ${v}_{y} \left(t\right) = 0$, so when $g t = {v}_{0} \sin \left(\theta\right)$.

${t}_{m} = \frac{{v}_{0} \sin \left(\theta\right)}{g} = \frac{2 \sin \left(\frac{\pi}{8}\right)}{g}$.

Its $x$ distance from the starting point at this time will be $x \left({t}_{m}\right) = \frac{{v}_{0} \cos \left(\theta\right) \cdot {v}_{0} \sin \left(\theta\right)}{g} = \frac{4 \sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right)}{g}$.

Its $y$ distance from the starting point at this time will be $y \left({t}_{m}\right) = - \frac{1}{2} g {\left(\frac{{v}_{0} \sin \left(\theta\right)}{g}\right)}^{2} + {v}_{0} \sin \left(\theta\right) \left(\frac{{v}_{0} \sin \left(\theta\right)}{g}\right) = \frac{{v}_{0}^{2} {\sin}^{2} \left(\theta\right)}{2 g} = \frac{2 {\sin}^{2} \left(\frac{\pi}{8}\right)}{g}$.

So the total distance $d$ from the starting point is given by the Pythagorean Theorem on the distances, i.e. ${d}^{2} = {x}^{2} + {y}^{2}$

So after all the algebra, we get $d = \frac{2 \sin \left(\frac{\pi}{8}\right) \left(4 {\cos}^{2} \left(\frac{\pi}{8}\right) + {\sin}^{2} \left(\frac{\pi}{8}\right)\right)}{g}$.

Nov 23, 2016

The distance form the starting point, $d \approx 0.15 m$

#### Explanation:

The equation for the y coordinate is:

y(t) = ((-4.9" m")/"s"^2)t^2 + ((2" m")/s")sin(pi/8)t

The equation for the x coordinate is:

x(t) = ((2" m")/s")cos(pi/8)t

The time that the projectile is at is maximum height is at the axis of symmetry of the parabola:

$t = - \frac{b}{2 a}$

$t = - \frac{2 \sin \left(\frac{\pi}{8}\right)}{- 9.8} s$

$t \approx 0.08 \text{ s}$

$y \left(0.08\right) \approx 0.03 m$

$x \left(0.08\right) \approx 0.15 m$

The distance, d, at the above time is:

$d = \sqrt{{\left(0.03\right)}^{2} + {\left(0.15\right)}^{2}}$

$d \approx 0.15 m$