# Question #eb793

##### 1 Answer

#### Answer:

#### Explanation:

The idea here is that you can use the distance traveled by the object in *its last 2 seconds* of free fall to determine its **velocity** as it enters this final stage in its fall.

So, let's assume that the object has a velocity **right before** entering the final

#h_"f" = v_1 * t_"f" + 1/2 * g * t_"f"^2" "# , where

**2 seconds** needed to travel these

Plug in your values and solve for

#v_1 = (h_"f" - 1/2 * g * t_"f"^2)/t_"f"#

#v_1 = ("80 m" - 1/2 * 9.8"m"color(red)(cancel(color(black)("s"^(-2)))) * 2^2 color(red)(cancel(color(black)("s"^(-2)))))/("2 s") = "30.2 m/s"#

Now, let's say that the **initial height traveled by the object** is equal to

#h_0 = h_"i" + h_"f"#

Here **before** reaching the

Since the object starts from rest, you can say that

#v_1^2 = overbrace(v_0^2)^(color(blue)(= "0 m/s")) + 2 * g * h_"i"#

Plug in your values and solve for

#h_"i" = v_1^2/(2 * g)#

#h_"i" = (30.2^2 "m"^color(red)(cancel(color(black)(2)))color(red)(cancel(color(black)("s"^(-2)))))/(2 * 9.8color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-2))))) = "46.53 m"#

This means that the object fell from an initial height

#h_0 = "46.53 m" + "80 m" = "126.53 m"#

Now, the answer *should* be rounded off to one sig fig, but I'll leave it rounded to three sig figs

#h_0 = color(green)("127 m")#