# A projectile is shot from the ground at an angle of ( pi)/3  and a speed of 6 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?

Oct 12, 2016

Let the velocity of projection of the object be u with angle of projection $\alpha$ with the horizontal direction.
The vertical component of the velocity of projection is $u \sin \alpha$ and the horizontal component is $u \cos \alpha$

Now if the time of flight be T then the object will return to the ground after T sec and during this T sec its total vertical displacement h will be zero. So applying the equation of motion under gravity we can write
$h = u \sin \alpha \times T + \frac{1}{2} g {T}^{2}$ $\implies 0 = u x T - \frac{1}{2} \times g \times {T}^{2}$
where $g = \text{acceleration due to gravity}$

$\therefore T = \frac{2 u \sin \alpha}{g}$
The horizontal displacement during this T sec is $R = u \cos \alpha \times T = \frac{{u}^{2} \sin \left(2 \alpha\right)}{g}$

The time t to reach at the peak is half of time of flight (T)
So $t = \frac{1}{2} \cdot T = \frac{u \sin \alpha}{g}$
The horizontal displacement during time t is

${d}_{h} = \frac{1}{2} \times \frac{{u}^{2} \sin \left(2 \alpha\right)}{g}$

If H is the maximum height then

${0}^{2} = {u}^{2} {\sin}^{2} \left(\alpha\right) - 2 \cdot g \cdot H$

$\therefore H = \frac{{u}^{2} {\sin}^{2} \left(\alpha\right)}{2 \cdot g}$

So the distance of the object from the point of projection when it is on the peak is given by

$D = \sqrt{{d}_{h}^{2} + {H}^{2}}$

=sqrt((u^2sin^2(2alpha))/(2g)^2+(u^4sin^4alpha)/(2g)^2

=u/(2g)sqrt((sin^2(2alpha))+(u^2sin^4alpha)

=6/(2*9.8)sqrt(sin^2((2pi)/3)+6^2sin^4(pi/3)

$= \frac{3}{9.8} \sqrt{\frac{3}{4} + \frac{36 \cdot 9}{16}}$

$= \frac{3}{9.8} \sqrt{\frac{3}{4} + \frac{81}{4}}$

$= \frac{3}{9.8} \sqrt{21} m \approx 1.4 m$