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?​

Answer 1

Here Is Your Answer..

Given

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?

Solution

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

Therefore,

4.27s

Hope It Helps!

$\pink{ \rule{19cm}{0.2cm}}$

Answer 2

Solution

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

Therefore,

4.27s .