Question

A red ball is dropped from rest at a height of 7.80 m. A blue ball at a height of 11.7 m is thrown down at the same instant at 9.00 m/s. How long does it take the blue ball to catch up with the red ball?

Answer 1

It takes the blue ball 0.4333 seconds to catch up with the red ball.

Step-by-step explanation:

We can use the equation

`y=v_ot-(1)/(2)gt^2`

Note
`y` is the change in position on the y-axis. So there is an initial and final component.

We can use that equation to find out at what time
`t`, the 2 balls will have traveled the same distance.

Let the red ball be object 1 and let the blue ball be object 2.

`y_(o1) -(v_(o1)t-(1)/(2)gt^2)=y_(o2) -(v_(o1)t-(1)/(2)gt^2)`

Solving for
`t` gives us

`t=-(y_(o2) -y_(o1))/(v_(o1) -v_(o2))`

We are given

`y_(o1)=7.8`

`y_(o2)=11.7`

`v_(o1)=0`

`v_(o2)=9`

Substituting our values into the equation gives us

`t=-(11.7-7.8)/(0-9)`

`t=-(3.9)/(0-9)`

`t=-(3.9)/(-9)`

`t=0.4333`