# A projectile is shot from the ground at an angle of pi/6  and a speed of 1 m/s. When the projectile is at its maximum height, what will its distance, factoring in height and horizontal distance, from the starting point be?

May 12, 2016

H=0.012m and ${d}_{h} = 0.044 m$

#### Explanation:

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 dosplacement h will be zero. So applying the equation of motion under gravity we can write

$h = u \sin \alpha \times T + \frac{1}{2} g {T}^{2}$
$\implies 0 = u \sin \alpha \times T - \frac{1}{2} \times g \times {T}^{2}$
where g= "acceleration" "dueto"" gravity”
$\therefore T = \frac{2 u \sin \alpha}{g}$
The horizontal displacement during this T sec is $R = u \cos \alpha \times T$
$R = \frac{{u}^{2} \sin \left(2 \alpha\right)}{g}$
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 ${d}_{h} = \frac{1}{2} \times \frac{{u}^{2} \sin \left(2 \alpha\right)}{g}$
By the problem
u=1m/s;g=9.8m/s^2 and alpha= pi/6
${d}_{h} = \frac{1}{2} \times \left({1}^{2} \sin \frac{2 \cdot \frac{\pi}{6}}{9.8}\right) = 0.044 m$
If H is the maximum height then
${0}^{2} = {u}^{2} {\sin}^{2} \left(\alpha\right) - 2 \cdot g \cdot H$
$\therefore H = \frac{{u}^{2} {\sin}^{2} \left(\alpha\right)}{2 \cdot g} = \frac{{1}^{2} {\sin}^{2} \left(\frac{\pi}{6}\right)}{2 \cdot 9.8} = 0.012 m$

So the distance of the object from the point of projection when it is on the peak
Is given by
$D = \sqrt{{d}_{h}^{2} + {H}^{2}} = \sqrt{{\left(0.044\right)}^{2} + \left({0.012}^{2}\right)} = 0.045 m$