6. You throw a golf ball at a 45 degree angle at 10 meters per second. a. How far away does it land? Assume level ground, no wind and no air resistance. b. How long does the flight take?

This question gives us the 45-degree angular velocity of a golf ball. Our first step is to split that angular vector into the ball's upwards momentum (which fights gravity) and its forward momentum (which theoretically fights air resistance, but we're ignoring that for the moment).
The upwards and forwards momentum are perpendicular, forming a right triangle with the angular momentum as its hypotenuse. 45 degrees is, fortunately, an easy number to work with. Rather than mess around with trigonometry, sines, and cosines, we can simply use common 45/45/90 isosceles right triangle ratio to determine that the sides are in a ratio of 1 : 1 : sqrt(2), and thus that the upwards and forward velocities are both 10 / sqrt(2) = 5 sqrt(2) = ~7.07 m/s.
As I previously mentioned, gravity pulls down on the ball's upward momentum. The gravitational constant on earth is 9.8 meters per second per second. That means the golf ball will take 7.07 / 9.8 = ~0.72 seconds to reach the peak of its arc, where is has zero vertical velocity, and to start accelerating back downwards. Moreover, because resistance-less ballistic arcs are symmetrical, the ball takes exactly the same amount of time to return to ground. Therefore the total flight time is ~1.44 seconds.
Once we have the total flight time, we multiply by the forward momentum of 10 m/s to find that the ball lands 14.4 meters away from the starting point.