Question

Paul throws a rock straight up into the air on Planet A, reaching a maximum height of 45 feet at 3 seconds. Jim throws a rock straight up into the air on Planet B, reaching a maximum height of 38 feet at 4 seconds. If Paul's rock leaves his hand at a height of 5.4 feet and Jim's at 6 feet, what is the difference in heights of the two rocks 6 seconds after they are thrown?

Answer 1

The difference in heights of the two rocks 6 seconds after they are thrown is 0.6 feet.

Step-by-step explanation:

To find the difference in heights of the two rocks 6 seconds after they are thrown, we need to first calculate the positions of the rocks at this time. Let's start with Paul's rock on Planet A. We know that the maximum height reached by Paul's rock is 45 feet, which it reaches at 3 seconds.

This means that at 6 seconds, the rock would have already fallen back to the ground. Since the rock is thrown straight up and comes back down, the height at 6 seconds would be equal to the initial height of the throw, which is 5.4 feet.

Now let's move on to Jim's rock on Planet B. We know that the maximum height reached by Jim's rock is 38 feet, which it reaches in 4 seconds.

Since the rock is thrown straight up and comes back down, the height at 6 seconds would be equal to the initial height of the throw, which is 6 feet.

Therefore, the difference in heights of the two rocks 6 seconds after they are thrown is 5.4 - 6 = -0.6 feet. This means that Jim's rock is 0.6 feet higher than Paul's rock at this time.