Question

Question #eb793

Answer 1

`"127 m"`

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 `v_1` right before entering the final `"80 m"` of its fall. This means that you can say

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

`h_"f"` - the final `80` meters
`t_"f"` - the 2 seconds needed to travel these `80` meters

Plug in your values and solve for `v_1` to get

`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_"i"`. In this case, you an say that

`h_0 = h_"i" + h_"f"`

Here `h_"i"` represents the distance it traveled before reaching the `80` it traveled in the last 2 seconds of flight.

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"` to get

`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` of

`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")`