# A projectile is shot from the ground at an angle of #pi/4 # and a speed of #16 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?

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The horizontal distance at the maximum height ((d_{\text{max}})) can be found using the formula: [d_{\text{max}} = \frac{\text{initial velocity}^2 \times \sin^2(\text{launch angle})}{2 \times \text{acceleration due to gravity}}] Substitute the given values: [d_{\text{max}} = \frac{16^2 \times \sin^2(\frac{\pi}{4})}{2 \times 9.8}] Calculate the result.

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