Science, Tech, Math › Science Free Falling Body Share Flipboard Email Print C.J. Burton, Getty Images Science Physics Physics Laws, Concepts, and Principles Quantum Physics Important Physicists Thermodynamics Cosmology & Astrophysics Chemistry Biology Geology Astronomy Weather & Climate By Andrew Zimmerman Jones Math and Physics Expert M.S., Mathematics Education, Indiana University B.A., Physics, Wabash College Andrew Zimmerman Jones is a science writer, educator, and researcher. He is the co-author of "String Theory for Dummies." our editorial process Andrew Zimmerman Jones Updated November 19, 2019 One of the most common sorts of problems that a beginning physics student will encounter is to analyze the motion of a free-falling body. It's helpful to look at the various ways these sorts of problems can be approached. The following problem was presented on our long-gone Physics Forum by a person with the somewhat unsettling pseudonym "c4iscool": A 10kg block being held at rest above the ground is released. The block begins to fall under only the effect of gravity. At the instant that the block is 2.0 meters above the ground, the speed of the block is 2.5 meters per second. At what height was the block released? Begin by defining your variables: y0 - initial height, unknown (what we're trying to solve for)v0 = 0 (initial velocity is 0 since we know it begins at rest)y = 2.0 m/sv = 2.5 m/s (velocity at 2.0 meters above ground)m = 10 kgg = 9.8 m/s2 (acceleration due to gravity) Looking at the variables, we see a couple of things that we could do. We can use conservation of energy or we could apply one-dimensional kinematics. Method One: Conservation of Energy This motion exhibits conservation of energy, so you can approach the problem that way. To do this, we'll have to be familiar with three other variables: U = mgy (gravitational potential energy)K = 0.5mv2 (kinetic energy)E = K + U (total classical energy) We can then apply this information to get the total energy when the block is released and the total energy at the 2.0-meter above-the-ground point. Since the initial velocity is 0, there is no kinetic energy there, as the equation shows E 0 = K 0 + U 0 = 0 + mgy 0 = mgy 0E = K + U = 0.5mv2 + mgyby setting them equal to each other, we get:mgy0 = 0.5mv2 + mgyand by isolating y0 (i.e. dividing everything by mg) we get:y0 = 0.5v2 / g + y Notice that the equation we get for y0 doesn't include mass at all. It doesn't matter if the block of wood weighs 10 kg or 1,000,000 kg, we will get the same answer to this problem. Now we take the last equation and just plug our values in for the variables to get the solution: y0 = 0.5 * (2.5 m/s)2 / (9.8 m/s2) + 2.0 m = 2.3 m This is an approximate solution since we are only using two significant figures in this problem. Method Two: One-Dimensional Kinematics Looking over the variables we know and the kinematics equation for a one-dimensional situation, one thing to notice is that we have no knowledge of the time involved in the drop. So we have to have an equation without time. Fortunately, we have one (although I'll replace the x with y since we're dealing with vertical motion and a with g since our acceleration is gravity): v 2 = v 0 2+ 2 g( x - x 0) First, we know that v0 = 0. Second, we have to keep in mind our coordinate system (unlike the energy example). In this case, up is positive, so g is in the negative direction. v2 = 2g(y - y0)v2 / 2g = y - y0y0 = -0.5 v2 / g + y Notice that this is exactly the same equation that we ended up within the conservation of energy method. It looks different because one term is negative, but since g is now negative, those negatives will cancel and yield the exact same answer: 2.3 m. Bonus Method: Deductive Reasoning This won't give you the solution, but it will allow you to get a rough estimate of what to expect. More importantly, it allows you to answer the fundamental question that you should ask yourself when you get done with a physics problem: Does my solution make sense? The acceleration due to gravity is 9.8 m/s2. This means that after falling for 1 second, an object will be moving at 9.8 m/s. In the above problem, the object is moving at only 2.5 m/s after having been dropped from rest. Therefore, when it reaches 2.0 m in height, we know that it hasn't fallen very fall at all. Our solution for the drop height, 2.3 m, shows exactly this; it had fallen only 0.3 m. The calculated solution does make sense in this case.