Question

An object is dropped from the top of a tall building. At 2 seconds, it is 64 feet from the top of the building. At 4 seconds, it is 256 feet from the top of the building. What is the average rate the object was traveling in the interval between 2 and 4 seconds?

Answer 1

Answer:

96.21 ft/s

Explanation:

To solve this, you only need to use one expression which is:

Vf² = Vo² + 2gh

g = 9.8 m/s²

However, this exercise is talking in feet, so convert the gravity to feet first:

g = 9.8 * 3.28 = 32.15 ft/s²

Vo is zero, because it's a free fall and in free fall the innitial speed is always zero. With this, let's calculate the speed at 2 seconds, with a height of 64 ft, and then with the 256 ft:

V1 = √2*32.15*64

V1 = 64.15 ft/s

V2 = √2*32.15*256

V2 = 128.3 ft/s

So the average rate is:

V = 128.3 + 64.15 / 2

V = 96.22 ft/s