One Dimensional Kinematics ("Two Rocks")

Question

A rock is held steady over a cliff and dropped. 1 seconds later, another rock is thrown straight down at a speed of 11.3 m/s, and hits the first rock. How far have the rocks dropped before they collide? How long is the first rock in the air before it gets hit by the second rock?

What is the displacement of the rocks when they collide? __________ (Hint: be careful of sign -- the rocks drop.)

The first rock is in the air for __________ before it is hit.

Answer 1

One of Newton's Kinematic Equations of Motion is:

x = v_it + 1/2at² where x is the distance traveled, v_i is the initial velocity, a is the acceleration (in this case that is -9.8 m/s² (also known as "g"), and t is the time traveled.

If "t" represents the time traveled from the time rock #2 is dropped until the collision, then the time traveled for rock #1 will be "t + 1". And, since rock #1 is dropped making its initial velocity = 0, then the distance rock #1 travels is:

x = (0)(t + 1) + 1/2(-9.8)(t + 1)² = -4.9(t² + 2t + 1) = -4.9t² - 9.8t - 4.9

The distance rock #2 travels is:

x = -11.3t + 1/2(-9.8)t² = -11.3t - 4.9t²

And since the distances must be equal when the rocks collide:

-4.9t² - 9.8t - 4.9 = -11.3t - 4.9t²

-9.8t - 4.9 = -11.3t

-4.9 = -1.5t

t = 3.267 s

So the distance they traveled can be calculated by plugging the 3.267 s back into either equation:

x = -11.3(3.267) - 4.9(3.267)² = -89.2 m or 89.2 m below where they began

The time the first rock was in the air is t + 1 = 3.267 + 1 = 4.267 s = 4.27 s