# A projectile is shot from the ground at an angle of pi/4  and a speed of 15 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?

May 5, 2017

The distance is $= 11.5 m$

#### Explanation:

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

initial velocity is ${u}_{y} = v \sin \theta = 15 \cdot \sin \left(\frac{1}{4} \pi\right)$

Acceleration is $a = - g$

At the maximum height, $v = 0$

We apply the equation of motion

$v = u + a t$

to calculate the time to reach the greatest height

$0 = 15 \sin \left(\frac{1}{4} \pi\right) - g \cdot t$

$t = \frac{15}{g} \cdot \sin \left(\frac{1}{4} \pi\right)$

$= 1.08 s$

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

To find the horizontal distance, we apply the equation of motion

$s = {u}_{x} \cdot t$

$= 15 \cos \left(\frac{1}{4} \pi\right) \cdot 1.08$

$= 11.5 m$

May 5, 2017

Here in our case
$\theta = \frac{\pi}{4}$
$u = 15$m/s
$\therefore {u}_{{y}_{1}} = 15 \sin \left(\frac{\pi}{4}\right) = \frac{15}{\sqrt{2}}$
${u}_{x} = 15 \cos \left(\frac{\pi}{4}\right) = \frac{15}{\sqrt{2}}$
When the projectile will reach maximum height
Velocity in $y$ direction ${u}_{{y}_{2}}$$= 0$ we
$\therefore$ Applying equating of motion along $y$ direction we get
${\left({u}_{{y}_{2}}\right)}^{2} = {\left({u}_{{y}_{1}}\right)}^{2} + 2 \left(- g\right) h$
From this equation we get $h = 5.74 m$
Now time required to reach the highest point
$t = {u}_{{y}_{2}} / g$
$\therefore$ distance moved in $x$ ditection in this time
$d = {u}_{x}$$\times {u}_{{y}_{2}} / g$
$\therefore d =$11.5$m$
$\therefore$Distance from starting point of the projectile till it reaches the maximum height is given by

$l = \sqrt{{d}^{2} + {h}^{2}}$
$\therefore l = 12.8 m$