Begin by defining your variables: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 2.5m/sec. The block was initially released at a height of how many meters.

*y*_{0}- initial height, unknown (what we're trying to solve for)*v*_{0}= 0 (initial velocity is 0)*y*= 2.0 m/s*v*= 2.5 m/s (velocity at 2.0 meters above ground)*m*= 10 kg*g*= 9.8 m/s (acceleration due to gravity)

### 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*= 0.5*mv*^{2}(kinetic energy)*E*=*K*+*U*(total classical energy)

Notice that the equation we get forE_{0}=K_{0}+U_{0}= 0 +mgy_{0}=mgy_{0}

E=K+U= 0.5mv^{2}+mgyby setting them equal to each other, we get:

mgy_{0}= 0.5mv^{2}+mgyand by isolating y

_{0}(i.e. dividing everything bymg) we get:

y_{0}= 0.5v^{2}/ g +y

*y*

_{0}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:

This is an approximate solution, since we are only using two significant figures in this problem.

y_{0}= 0.5 * (2.5 m/s)^{2}/ (9.8 m/s^{2}) + 2.0 m = 2.3 m

### 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):

First, we know thatv^{2}=v_{0}^{2}+ 2g(x-x_{0})

*v*

_{0}= 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.

Notice that this is

v^{2}= 2g(y-y_{0})v^{2}/ 2g=y-y_{0}y_{0}= -0.5v^{2}/g+y

*exactly*the same equation that we ended up with in 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/s^{2}. 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.