Horizontal distance traveled by a projectile fired from the ground can be modeled by $\displaystyle d = \frac {v_0^2\sin{2\theta}}{g}$, where $\displaystyle v_0$ is the initial velocity, $\displaystyle \theta$ is the launch angle in degrees, and "g" is acceleration due to gravity. (This model neglects air resistance and assumes a flat surface.)

a) A baseball is hit at an initial height of 5.5 feet with initial velocity of 95 ft/s at 42 degrees to the ground. Will the ball clear a 5 ft fence 310 feet from home plate?

I used a calculator to solve this one. I plugged in$\displaystyle y = \frac {95^2\sin{2x}}{9.8} +5.5$and find that when "x" is 42, the value of "y" is 921.3735. Since the "y" value is horizontal distance,the distance from home plate is 921.3735. My question is how do I know that the ball cleared the5 ft fence 310 ft away from home plate, since the "y" value is only the horizontal distance and not vertical distance?

b) What angle will maximize the horizontal distance traveled and why?

I find out that the angle that will maximize horizontal distance is 45 degrees, which is true in physics from what I've learned. At 45 degrees the baseballl will travel 926.42 feet. My question is why is 45 the angle that maximizes horizontal distance?

c) Will the ball clear the fence if it at the angle from part (b) above?

Just like in part (a),how do I know that the ball cleared the5 ft fence 310 ft away from home plate? I know it traveled926.42 feet horizontally, but I don't know if it cleared the 5 foot fence.