Sandbag Shadow Problem - Please Help!!

[wismadison2010]

Okay so....
A sandbag is dropped from a balloon at a height of 60 meters when the angle of elevation to the sun is 30 degrees. Find the rate at which the shadow of the sandbag is traveling along the ground when the sandbag is at a height of 35 meters.

Position Function: s(t) = 60-4.9t^2

Any help would be much appreciated! I'm thinking proportional but I'm really not sure...befuddled.

 

[stapel]

So you have a right triangle, formed by the height line (s(t) = 60 - 4.9t[sup:tg7bqwig]2[/sup:tg7bqwig]), the ground line (between the shadow and the base of the height line), and the line of sight. You are given that the base angle is thirty degrees, and you know that the relationship between the height and the ground distance (being the "opposite" and "adjacent" sides of the thirty-degree angle) is a certain fixed ratio.

See where this leads.... :wink:

 

[galactus]

\(\displaystyle s(t)=60-4.9t^{2}\)

\(\displaystyle s'(t)=-9.8t\)

\(\displaystyle s=35=60-4.9t^{2}\)

\(\displaystyle t=\frac{5}{\sqrt{4.9}}=\frac{5\sqrt{10}}{7}\approx 2.26\)

\(\displaystyle tan(30)=\frac{1}{\sqrt{3}}=\frac{s(t)}{x(t)}\)

\(\displaystyle x(t)=\sqrt{3}s(t)\)

\(\displaystyle \frac{dx}{dt}=\sqrt{3}\frac{ds}{dt}\)

\(\displaystyle =\sqrt{3}(-9.8)\frac{5}{\sqrt{4.9}}= -38.34 \;\ m/sec\)

 

[soroban]

Hello, wismadison2010!

A sandbag is dropped from a balloon at a height of 60 meters when the angle of elevation to the sun is 30 degrees.
Find the rate at which the shadow of the sandbag is traveling along the ground when the sandbag is at a height of 35 meters.

Position Function: .\(\displaystyle s(t) \:=\: 60-4.9t^2\)

    S ^
      |   *
      |       *
      |           *
 s(t)  |               *
      |                   *
      |                  30d  *
    B * - - - - - - - - - - - - - * A
                   x

\(\displaystyle \text{The sandbag is at }S\)
. . \(\displaystyle \text{Its height is: }\:SB \:=\:s(t) \:=\:60-4.9t^2\)
\(\displaystyle \text{Its shadow is at }A.\)
. . \(\displaystyle \text{Let }\,x \,=\,AB\)

\(\displaystyle \text{In the right triangle: }\,\tan30^o \:=\:\frac{s(t)}{x} \quad\Rightarrow\quad \frac{1}{\sqrt{3}} \:=\:\frac{s(t)}{x}\)

\(\displaystyle \text{Hence: }\:x \:=\:\sqrt{3}\cdot s(t) \:=\:\sqrt{3}(60 - 4.9t^2)\)

\(\displaystyle \text{Differentiate with respect to time: }\:\frac{dx}{dt} \;=\;\sqrt{3}(-9.8t) \;=\;-9.8\sqrt{3}\,t\) .[1]


\(\displaystyle \text{When }s = 35\text{, we have: }\:60-4.9t^2 \:=\:35 \quad\Rightarrow\quad 4.9t^2 \:=\:25\)

. . \(\displaystyle t^2\:=\:\frac{25}{4.9} \quad\Rightarrow\quad t^2 \:=\:\frac{250}{49} \quad\Rightarrow\quad t \:=\:\sqrt{\frac{250}{49}} \:=\:\frac{5\sqrt{10}}{7}\)


Substitute into [1]:

. . \(\displaystyle \frac{dx}{dt} \;=\;-9.8\sqrt{3}\left(\frac{5\sqrt{10}}{7}\right) \;=\;-7\sqrt{30}\)

\(\displaystyle \text{The shadow is moving at about }38.3\text{ m/sec}\)


Edit: corrected a silly error.
. . . Thanks for the heads-up, Cody!
.