A projectile is shot from the ground at an angle of (5 pi)/12  and a speed of 4/9 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?

Jan 17, 2016

$25.51 m m$.

Explanation:

In the vertical (y) direction, there is constant acceleration under gravity, so the equations of motion for constant linear acceleration are valid.
We may use them to find the time taken to reach maximum height
Note that initial velocity may be resolved into horizontal and vertical components, and vertical velocity at maximum height is zero at maximum height .

$\therefore v = u + a t \implies t = \frac{v - u}{a}$

$= \frac{0 - \frac{4}{9} \sin \left(\frac{5 \pi}{12}\right)}{-} 9.8 = 0.09856 s$.

In the horizontal (x) direction, there is no acceleration (constant velocity) so the definition of velocity as the rate of change in displacement is valid.

$\therefore {s}_{x} = {v}_{x} t$

$= \frac{4}{9} \cos \left(\frac{5 \pi}{12}\right) \times 0.09856$

$= 0.02551 m$.