Question

A ladder of length $17meters$ rests with one end against a vertical wall and the other on the level ground. if the lower end slipes away at the rate of $1m/sec$, then, when it is $8m$ away from the well, its upper end is coming down at the rate of

• A. $\frac{5}{8}m/sec$.
• B. $\frac{8}{15}m/sec$.
• C. $\frac{5\pi }{8}m/sec$.
• D. $\frac{15}{8}m/sec$.

Answer 1

The correct option is B $\frac{8}{15}m/sec$.

$\begin{array}{c}{x}^2+y^2={\left(17\right)}^2\\ 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\\ when,x=8y=\sqrt{289-64}=15\\ x\frac{dx}{dt}+y\frac{dy}{dt}=0\\ 8\times 1+15\frac{dy}{dt}=0\left[\frac{dx}{dt}=1m/s\right]given\\ \frac{dy}{dt}=\frac{-8}{15}\end{array}$

$-Ve$ sign shown that $y$ is decreasing.

Option $B$ is correct.