Question

A stone projected upwards with a velocity u reaches two points P and Q separated by a distance h with velocities u/2 and u/3. The maximum height reached by it is

• A. $\frac{9h}{5}$
• B. $\frac{18h}{5}$
• C. $\frac{36h}{5}$
• D. $\frac{72h}{5}$

Answer 1

Answer: (C)

At $ P, \left(\frac {u}{2}\right)^2 = u^2 - 2gs_1 - (1)$

At $ Q, \left(\frac {u}{3}\right)^2 = u^2 - 2gs_2 - (2)$

Maximum height $= \frac {u^2}{2g} - (3)$

$eq^n (2) - eq^n (1) \Rightarrow -2g(s_2 - s_1)$ = $\frac {u^2}{9} - \frac {u^2}{4}$

$\Rightarrow +2gh = \frac {5u^2}{36}$

$\Rightarrow \frac {36}{5}h = \frac {u^2}{2g} \rightarrow $ from $eq^n (3)$

Max ht $ = \frac {36h}{5}$