Question

A body of mass $m$ is dropped and another body of mass $M$ is projected vertically up with speed $u$ simultaneously from the top of a tower of height $H$. If the body reaches the highest point before the dropped body reaches the ground, then maximum height raised by the centre of mass of the system from ground is :

• A. $H+\frac{u^2}{2g}$
• B. $\frac{u^2}{2g}$
• C. $H+\frac{1}{2g}(\frac{Mu}{m+M}{)}^2$
• D. $H+\frac{1}{2g}(\frac{mu}{m+M}{)}^2$

Answer 1

The correct option is C $H+\frac{1}{2g}(\frac{Mu}{m+M}{)}^2$

At time $t=0$ sec

Velocity of $M=u$

Velocity of $m=0$

Velocity of COM$\left(v_c\right)$ = $\frac{M\times u+m\times 0}{M+m}$

Maximum height attained in vertical projectile is: $h_{max}=\frac{u^2}{2g}$

Maximum height attained by COM = $H+\frac{v_c^2}{2g}$

= $H+\frac{(\frac{Mu}{m+M}{)}^2}{2g}$