1D Kinematics Legacy Problem #26 Guided Solution
Problem*
According to Guinness, the tallest man to have ever lived was Robert Pershing Wadlow of Alton, Illinois. He was last measured in 1940 to be 2.72 meters tall (8 feet, 11 inches). Determine the speed which a quarter would have reached before contact with the ground if dropped from rest from the top of his head.
Audio Guided Solution
One of the critical skills that a beginning student in physics must develop is the skill of carefully reading a problem, extracting numerical information from the problem, either that which is explicitly stated or that which is implied, in an effort to generate enough information to solve for the unknown. For instance, here we have an object, which is a quarter, which is being dropped from the top of a tall person's head. We're told the person's 2.72 meters tall, so the object's going to fall to the ground. We're told it's dropped from rest, so that tells us that the V original is 0 meters per second. That's implied information. And it's a 2.72 distance drop, so we would say D equal negative 2.72 meters. And then the final bit of information that you need to identify here is this quarter's going to free fall to the ground. We're sort of making that an assumption, but it's a very good assumption. So we could say A equal negative 9.8 meters per second per second. So we know three bits of information. We're asked for a fourth bit of information, and that is, we're asked to find the speed at which it hits the ground. That is, find Vf. So if I'm doing this right, I would be writing things down. I'd be writing this down. I'd say D equal negative 2.72 meters. I would write down V0 equals 0 meters per second. A equal negative 9.8 meters per second per second. And then I would start to look for my equations, and in the overview page, there's a set of problems. What you'll notice is there's a discussion of four kinematic equations, and you're looking for the one that has within it the V0, the D, the A, and the Vf. And the one of choice is probably the one that goes Vf squared equals V0 squared plus 2Ad. And since V0 is 0, that first term right side cancels out. The equation simplifies to Vf squared equals 2Ad. You're looking for Vf. So plug in your negative 9.8 and plug in your negative 2.72 meters on the right side for the A and the D. Evaluate what that whole right side is, and then take the square root of each side, and you have your Vf. And that's all there is to it, this problem.
Solution
7.30 m/s
Habbits of an Effective Problem Solver
- Read the problem carefully and develop a mental picture of the physical situation. If necessary, sketch a simple diagram of the physical situation to help you visualize it.
- Identify the known and unknown quantities in an organized manner. Equate given values to the symbols used to represent the corresponding quantity - e.g., \(v_o = \units{0}{\unitfrac{m}{s}}\); \(a = \units{4.2}{\unitfrac{m}{s^2}}\); \(v_f = \units{22.9}{\unitfrac{m}{s}}\); \(d = \colorbox{gray}{Unknown}\).
- Use physics formulas and conceptual reasoning to plot a strategy for solving for the unknown quantity.
- Identify the appropriate formula(s) to use.
- Perform substitutions and algebraic manipulations in order to solve for the unknown quantity.
Read About It!
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