# A projectile is shot from the ground at an angle of pi/4  and a speed of 2 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 11, 2016

The horizontal distance is $= \frac{2}{g} = 0.2 m$

#### Explanation:

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

The vertical component of the velocity $= u \sin \theta = 2 \cdot \sin \left(\frac{\pi}{4}\right)$

$= 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2}$

as $\cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

The horizontal component of the velocity $= u \sin \theta = 2 \cdot \cos \left(\frac{\pi}{4}\right)$

$= 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2}$

as $\sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

We use the equation, $v = u + a t$

So, $0 = \sqrt{2} - >$

$t = \frac{\sqrt{2}}{g}$

To find the height, we use ${v}^{2} = {u}^{2} + 2 a s$

$0 = 2 - 2 g h$

$h = \frac{2}{2 g} = \frac{1}{g}$

The horizontal distance is $= {u}_{0} \cdot t = \sqrt{2} \cdot \frac{\sqrt{2}}{g} = \frac{2}{g}$

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