5. You throw a golf ball straight up at 10 meters per second. a. What is the velocity when you catch the golf ball after its return trip? b. How long did the golf ball’s total trip take (in seconds)? c. What is the total distance the golf ball traveled (up and down)?

When you throw a golf ball straight up, assuming that wind and air resistance are irrelevant, it is pulled back down by gravity, which decelerates it by 9.8 meters/second every second. That deceleration is constant through its rise and fall, creating a parabolic arc when considering its speed graphically.
The answer to part a) is 10 m/s. No calculation is necessary. Acceleration curves are symmetrical; in the absence of air resistance, a rising and falling object will always land at the same speed at which it was launched.
To answer part b), we can do a simple calculation involving the gravitational constant. Gravity robs the golf ball of speed at its constant rate, so it will take 10 / 9.8 = ~1.02 seconds for the golf ball's velocity to reach zero at the apex of its flight. Because rise and fall are symmetrical, we do not need calculations to know that it will take another 1.02 seconds for the upward motion to reverse, reaching your hand once again. The total time the golf ball travels is 2 * ~1.02 = ~2.04 seconds
Finally, part c) is once again fairly simple. Because acceleration is constant, the average velocity of a fall is always equal to exactly half the final velocity. Because of the symmetry of physics, the acceleration upwards perfectly mirrors the fall back down. Since the golf ball started and ended its flight at 10 m/s, its average speed is 5 m/s, meaning it traveled a total distance of 5 * 2.04 = 10.08 meters.