Question #cb1b7

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 #"2 ms"^(-1)# and accelerate towards the water under the influence of gravity.

On the other hand, the cage will continue its descent at a constant velocity of #"2 m s"^(-1)#.

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

Let's assume that after the object falls, the cage reaches the water in #t_c# seconds. You know that the object reaches the water #10# seconds 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 the cage

and

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

This means that you have

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

Use equation #color(red)("(*)")# to find

#t_c = t_0 + 10#

Plug this into equation #color(purple)("(*)")# to get

#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_o#

#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")#