# Question #cb1b7

##### 1 Answer

#### Answer:

#### Explanation:

The key to this problem lies with understanding that the object falls with an **initial velocity** equal to the velocity of the cage.

The object will fall with an initial velocity of **accelerate** towards the water under the influence of gravity.

On the other hand, the cage will continue its descent at a **constant velocity** of

Another important thing to realize here is that both the cage and the object cover *the same distance*, which we'll label

Let's assume that after the object falls, the cage reaches the water in **before** the cage, so you can say that

#t_o = t_c - 10" " " "color(red)("(*)")#

So, you can write two equations

#h = v_o * t_c -># for thecage

and

#h = v_0 * t_o + 1/2 * g * t_o^2 -># for theobject

This means that you have

#v_0 * t_c = v_0 * t_o + 1/2 * g * t_o^2" " " "color(purple)("(*)")#

Use equation

#t_c = t_0 + 10#

Plug this into equation

#v_0 * (t_0 + 10) = v_0 * t_o + 1/2 * g * t_o^2#

This simplifies to

#color(red)(cancel(color(black)(v_0 * t_0))) + 10 * v_0 = color(red)(cancel(color(black)(v_0 * t_0))) + 1/2 * g * t_o^2#

#1/2 * g * t_o^2 = 10 * v_0#

Rearrange to find

#t_0^2 = (2 * 10 * v_0)/g implies t_0 = sqrt((2 * 10 * v_0)/g)#

Plug in your values to get

#t_o = sqrt( (2 * 10color(red)(cancel(color(black)("s"))) * 2 color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))/(9.81color(red)(cancel(color(black)("m")))"s"^(-2))) = "2.02 s"#

This means that

#t_c = "2.02 s" + "10 s" = "12.02 s"#

The height from which the object fell is equal to

#h = v_0 * t_c#

#h = "2 m"color(red)(cancel(color(black)("s"^(-1)))) * 12.02color(red)(cancel(color(black)("s"))) ~~ color(green)("24 m")#