Question

A ladder $10$ m long rests against a vertical wall with the lower end on the horizontal ground.

The lower end of the ladder is pulled along the ground away from the wall at the rate of $3$ cm/s.

The height of the upper end while it is descending at the rate of $4$ cm/s, is

• A. $4\sqrt{3}\mathrm{m}$
• B. $5\sqrt{3}\mathrm{m}$
• C. $5\sqrt{2}\mathrm{m}$
• D. $8\mathrm{m}$
• E. $6\mathrm{m}$

Answer 1

The correct option is E

$6\mathrm{m}$

Step1. Drawing the diagram according to the given data:

Given, A ladder 10 m long rests against a vertical wall with the lower end on the horizontal ground.

Therefore applying Pythagoras theorem.

$$\begin{gathered} {\mathrm{x}}^2+{\mathrm{y}}^2={10}^2 \\ {\mathrm{x}}^2+{\mathrm{y}}^2=100...\left(\mathrm{i}\right) \end{gathered}$$

Also, $\frac{\mathrm{dx}}{\mathrm{dt}}=3\frac{\mathrm{cm}}{\sec }$

And, $\frac{\mathrm{dy}}{\mathrm{dt}}=-4\frac{\mathrm{cm}}{\sec }$

Step2. Finding differentiation.

Differentiating equation with respect to time 't'.

$\begin{array}{rcl}\Rightarrow 2x\frac{dx}{dt}+2y\frac{dy}{dt}& =& 0\\ \Rightarrow x\frac{dx}{dt}+y\frac{dy}{dt}& =& 0\\ \Rightarrow 3x–4y& =& 0\\ \Rightarrow 3x& =& 4y\\ \Rightarrow x& =& \frac{4y}{3}\end{array}$

Step3. Finding height of the upper end.

Put value of ‘x’ in equation $\left(\mathrm{i}\right)$

$\begin{array}{rcl}& \Rightarrow & {\left(\frac{4\mathrm{y}}{3}\right)}^2+{\mathrm{y}}^2=100\\ & \Rightarrow & 16{\mathrm{y}}^2+9{\mathrm{y}}^2=900\\ \Rightarrow 25{\mathrm{y}}^2& =& 900\\ \Rightarrow {\mathrm{y}}^2& =& 36\\ \Rightarrow \mathrm{y}& =& 6\mathrm{m}\end{array}$

Hence, option (E) is the correct answer.