# Question eb793

Nov 15, 2015

$\text{127 m}$

#### 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 $\text{80 m}$ of its fall. This means that you can say

${h}_{\text{f" = v_1 * t_"f" + 1/2 * g * t_"f"^2" }}$, where

${h}_{\text{f}}$ - the final $80$ meters
${t}_{\text{f}}$ - 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 ${h}_{\text{i}}$. In this case, you an say that

${h}_{0} = {h}_{\text{i" + h_"f}}$

Here ${h}_{\text{i}}$ 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 ${h}_{\text{i}}$ to get

${h}_{\text{i}} = {v}_{1}^{2} / \left(2 \cdot g\right)$

${h}_{\text{i" = (30.2^2 "m"^color(red)(cancel(color(black)(2)))color(red)(cancel(color(black)("s"^(-2)))))/(2 * 9.8color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-2))))) = "46.53 m}}$

This means that the object fell from an initial height ${h}_{0}$ of

${h}_{0} = \text{46.53 m" + "80 m" = "126.53 m}$

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

${h}_{0} = \textcolor{g r e e n}{\text{127 m}}$