Question

A lamp is $50$ ft. above the ground. A ball is dropped from the same height from a point $30$ ft. away from the light pole. If ball falls a distance $s=16t^2$ ft. in $t$seconds, then the speed of the shadow of the ball moving along the ground $\frac{1}{2}$ s later is

• A. $-1500\text{ft/s}$
• B. $1500\text{ft/s}$
• C. $-1600\text{ft/s}$
• D. $1600\text{ft/s}$

Answer 1

The correct option is A

$-1500\text{ft/s}$

Explanation for the correct option:

Step 1: Draw the diagram according to the given information:

At time $t$ the ball drops $16t^2$ ft. distance.

So $y=50–16t^2$

Step 2: Finding the value of $\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathit{\theta }$

Point $A$ is the position of the falling ball at some time $t$.

So,$\frac{dy}{dt}=-32t$

From figure,

$\begin{array}{rcl}\tan \theta & =& \frac{AB}{BC}=\frac{y}{x}\\ & =& \frac{50}{(30+x)}\end{array}$

Step 3: Finding the speed of the shadow

$\begin{array}{rcl}y& =& \frac{50x}{(30+x)}\end{array}$

Differentiate it with respect to $x$

$\begin{array}{rcl}\frac{dy}{dt}& =& \left[\frac{1500}{{(30+x)}^2}\right]\frac{dx}{dt}\end{array}$

$\Rightarrow$$\begin{array}{rcl}\frac{dx}{dt}& =& -32t\times \frac{{(30+x)}^2}{1500}\end{array}$

At $t=\frac{1}{2}$, $\begin{array}{rcl}y& =& 50–4\\ & =& 46\end{array}$

Put $y=46$ in $y=\frac{50x}{(30+x)}$

$\Rightarrow$ $\begin{array}{rcl}46& =& \frac{50x}{(30+x)}\end{array}$

$\Rightarrow$$\begin{array}{rcl}1380+46x& =& 50x\end{array}$

$\Rightarrow$ $\begin{array}{rcl}4x& =& 1380\end{array}$

$\Rightarrow$ $\begin{array}{rcl}x& =& 345\end{array}$

$\begin{array}{rcl}\therefore \frac{dx}{dt}& =& -32t\times \frac{{(30+x)}^2}{1500}\\ & =& -\frac{16{(30+345)}^2}{1500}\\ & =& -1500\end{array}$

The speed of the shadow of the ball moving along the ground $\frac{1}{2}$s later is $\mathbf{-}\mathbf{1500}\mathbf{}\mathbf{feet}\mathbf{/}\mathbf{seconds}$.

Hence, option (A) is the correct answer.