4. You drop a 16-pound bowling ball off of a 50-foot structure. How fast is the bowling ball traveling when it strikes the ground?

This question is missing some basic assumptions, but based on the standard way these questions are typically phrased, I am assuming that the friction and air resistance are irrelevant.
Given that, the weight of the bowling ball is also irrelevant. Gravity on Earth is a constant that affects all objects equally, pulling at approximately 32.2 feet per second per second.
The acceleration is constant but that means the bowling ball's speed is not. The longer it falls, the faster it falls. To account for that discrepancy, the formula for calculating time falling involves the square root of the height:
Time to fall = sqrt ( 2 * height / 32.2 )
If we plug in our initial height into that formula and run it through our calculators, we get ~1.76 seconds until the ball hits the ground.
We can then plug that back into the gravitational constant to determine the bowling ball's final speed. The bowling ball gains 32.2 feet/second of speed for every second of acceleration, so, after 1.76 seconds of acceleration, it is traveling at,
32.2 ft / sec^2 * 1.76 sec = ~56.7 feet / sec
The bowling ball will be going about 56.7 feet per second when it strikes the ground.