# A projectile is shot from the ground at an angle of pi/6  and a speed of 9 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?

Apr 16, 2016

$\approx 4.04 m$

#### Explanation:

The velocity of projection of the projectile $u = 9 m {s}^{-} 1$
The angle of projection of the projectile $\alpha = \frac{\pi}{6}$
The horizontal component of velocity of projection
${u}_{h} = u \cos \alpha = 9 \times \cos \left(\frac{\pi}{6}\right) = 4.5 \sqrt{3} m {s}^{-} 1$
TheVertical component of velocity of projection
${u}_{v} = u \cos \alpha = 9 \times \sin \left(\frac{\pi}{6}\right) = 4.5 m {s}^{-} 1$


Assuming the ideal situation where gravitational pull is the only force acting on the body and no air resistance exists , we can easily proceed for various calculation using equation of motion under gravity. CALCULATION
Let the projectile reaches its maximum height H m after t s of its start.
The final vertical component of its velocity at maximum height will be zero
So we can write
$0 = u \sin \alpha - g \times t \implies t = \frac{u \sin \alpha}{g} = \frac{4.5}{10} = 0.45 s$ taking $g = 10 m {s}^{-} 2$

Again
${0}^{2} = {\left(u \sin \alpha\right)}^{2} - 2 \times g \times H$
$\implies H = {\left(u \sin \alpha\right)}^{2} / g = {\left(4.5\right)}^{2} / 10 = 2.025 m$

During $t = 1.25 s$ of its ascent its horizontal component will remain unaltered and that is why the horizontal displacement
$R = u \cos \alpha \times t = 4.5 \sqrt{3} \times 0.45 m = 3.5 m$

Hence the projectil's distance ( D ) from starting point to the maximum point of its ascent is given by
$D = \sqrt{{R}^{2} + {H}^{2}} = \sqrt{{3.5}^{2} + {\left(12.025\right)}^{2}} \approx 4.04 m$