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

The distance from the starting point is, then

Applying the equation of motion

Therefore,

Applying the equation of motion

The distance travelled in the horizontal direction is

The distance from the starting point is

To find the distance from the starting point when the projectile reaches its maximum height, you can use the following formula:

$\text{Horizontal distance} = \text{initial velocity} \times \text{time}$

First, find the time it takes for the projectile to reach its maximum height using the vertical motion equation:

$\text{Vertical velocity} = \text{initial vertical velocity} + (\text{acceleration due to gravity} \times \text{time})$

At maximum height, vertical velocity is 0:

$0 = \text{initial vertical velocity} + (\text{acceleration due to gravity} \times \text{time})$

$0 = 9 \sin(\frac{\pi}{3}) - 9.8t$

Solve for $t$:

$t = \frac{9 \sin(\frac{\pi}{3})}{9.8}$

Now, find the horizontal distance:

$\text{Horizontal distance} = 9 \cos(\frac{\pi}{3}) \times t$

$\text{Horizontal distance} = 9 \cos(\frac{\pi}{3}) \times \frac{9 \sin(\frac{\pi}{3})}{9.8}$

$\text{Horizontal distance} \approx 9 \times \frac{\sqrt{3}}{2} \times \frac{9 \times \frac{1}{2}}{9.8}$

$\text{Horizontal distance} \approx \frac{81 \sqrt{3}}{19.6}$

$\text{Horizontal distance} \approx 12.59 \, \text{meters}$