A player throws a ball from a height of 3.5 meters. The ball flies at a speed of 9.8 m/sec creating an angle of 30° parallel to the surface. Another player catches the ball 2.1 meters from the surface. What is the distance between those two players? Note: g = 9.8 m/sec²

$v_{0x} = v_0 \cos(\theta) = 9.8 \cos(30^\circ)$

$v_{0y} = v_0 \sin(\theta) = 9.8 \sin(30^\circ)$

$y = y_0 + v_{0y}t + \frac{1}{2}gt^2$

$2.1 = 3.5 + (9.8 \sin(30^\circ))t + \frac{1}{2}(-9.8)t^2$

$2.1 = 3.5 + (9.8 \times 0.5)t - 4.9t^2$

$2.1 = 3.5 + 4.9t - 4.9t^2$

$4.9t^2 - 4.9t + (2.1 - 3.5) = 0$

$4.9t^2 - 4.9t - 1.4 = 0$

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 4.9$, $b = -4.9$, and

$c = -1.4$:

$t = \frac{-(-4.9) \pm \sqrt{(-4.9)^2 - 4(4.9)(-1.4)}}{2(4.9)}$

$t = \frac{4.9 \pm \sqrt{24.01 + 27.44}}{9.8}$

$t = \frac{4.9 \pm \sqrt{51.45}}{9.8}$

$t = \frac{4.9 \pm 7.1728}{9.8}$

$t = \frac{4.9 + 7.1728}{9.8} = \frac{12.0728}{9.8} \approx 1.2319 \text{ s}$

$x = v_{0x}t$

$x = (9.8 \cos(30^\circ)) \times 1.2319$

$x = \left(9.8 \times \frac{\sqrt{3}}{2}\right) \times 1.2319$

$x \approx (9.8 \times 0.866) \times 1.2319$

$x \approx 8.487 \times 1.2319 \approx 10.45 \text{ m}$