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

Jan 14, 2016

$s \approx 0.099 m$

#### Explanation:

Step 1: Maximum height will be reached when $\sin \theta$ component of projectile's velocity becomes 0 (zero)

Use the formula
$v = u + g t$
Given $v = 0 , u = \frac{3}{2} . \sin \left(\frac{\pi}{6}\right)$; as $\theta = \frac{\pi}{6}$
Take $g = 9.81 \frac{m}{s} ^ 2$
Since gravity is acting against the velocity so it is deceleration and $-$ sign is used in front of g
$t = \frac{\frac{3}{2} . \sin \left(\frac{\pi}{6}\right)}{9.81} s$

Step 2:Distance traveled horizontally in time t is due to $\cos \theta$ component of the initial velocity of the projectile. Assuming zero air friction, and using the equation

$s = u t$
$s = \frac{3}{2} . \cos \left(\frac{\pi}{6}\right) \times \frac{\frac{3}{2} . \sin \left(\frac{\pi}{6}\right)}{9.81} m$