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

Answer:  

4.27s

Explanation:

If "t" represents the time traveled from the time rock 2 is dropped until the collision, then the time traveled for rock 1 = 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  

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

For 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

Now, the distance they traveled can be found 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

Answer 2

Let

First rocks time be x

Second rocks time be x+1

initial velocity=u=11.3m/s

Distance of both rocks be s1 and s2

$\\ \sf\longmapsto s=ut+\dfrac{1}{2}at^2$

Now

$\\ \sf\longmapsto s1=11.3x+5x^2$

$\\ \sf\longmapsto s2=11.3(x+1)+5(x+1)^2$

As both collide then

$\\ \sf\longmapsto s1=s2$

$\\ \sf\longmapsto 11.3x+5x^2=11.3x+11.3+5(x+1)^2$

$\\ \sf\longmapsto 5x^2=11.3+5x^2+10x+1$

$\\ \sf\longmapsto 10x+12.3=0$

$\\ \sf\longmapsto x=1.23s$

Displacement

$\\ \sf\longmapsto 11.3(1.23)=13.8m$

Done