# Kinematics: Free Falling Problem

This post will focus on utilizing the basic kinematics equations with one-dimensional motion. We will be analyzing two balls being dropped from a certain height, and how gravity affects these two objects.

Problem: Two students are on a balcony a distance of 50 meters above the street. One student throws a ball vertically downward at a speed of 3 m/s. At the same time, the other student throws a ball vertically upward at the same speed. Find the final velocity of both balls just before they strike the ground. The second part of the problem asks you to determine how far apart the 2 balls are 3.7 seconds after the students release them from their balcony.

Like any physics problem, our first step is to draw the problem out to help us conceptualize the problem and what the problem is asking us. We draw a balcony 50 meters high. Since we will label the up direction as positive, the ball that was thrown downward at an initial velocity of 3 m/s has an initial velocity of -3 m/s. The ball that was thrown upward at the same velocity has an initial velocity of +3 m/s.

Our next step is to write out the given variables in the problem. Since we are working with 2 balls, we will write 2 columns. For Ball A, we write all the information we can find about its actions (i.e. initial velocity, heights, acceleration). In the second column, we again write all the information we can find about its actions.

The problem is first asking what the final velocity of Ball A and Ball B right before they hit to ground. To do this, we have to utilize a kinematics equation that satisfies the information we are given. In this case

$v_{f}^{2} = v^{2}_{i}+2a\Delta X$

Inputting our variables, we find that the final velocity of both Ball A and Ball B is -31.4 m/s.  At first thought, you might say, but wait, one ball was thrown upward, the other downward.  How can they reach the same final velocity.  When Ball B is thrown upward at 3m/s, when it reaches the height that Ball A was dropped from, Ball B is now at a velocity of -3 m/s.  The same velocity as Ball A.

The second part of our problem is determining how far apart the two balls are from each other, 3.7 seconds after both balls were released.  To answer this question we have to use a kinematics equation that incorporates distance, initial velocity, time and gravity.  We must use the following equation to solve for distance:

$\Delta y=v_{i}t+\frac{1}{2}at^{^{2}}$

Since $\Delta y = y_{f} - y_{i}$ with a little rearrangement, our equation can read: $y_{f}=y_{i}+v_{i}t+\frac{1}{2}at^{^{2}}$

Filling in our known variables for Ball A, $y_{f} = -28.181$ meters. For Ball B $y_{f} = -5.981$ meters. Which makes our total distance between the two balls 22.2 meters.

The full answer to this problem can be viewed below.

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