Question

Two particles of the same mass are projected from the ground with the same speed $10\text{m/s}$ at an angle ${37}^{\circ }$ and ${53}^{\circ }$ with the horizontal respectively. Find the maximum height attained by their $COM$ (in metres). Take $(g=9.8{\text{m/s}}^2)$ and answer upto two decimal places.

• A. 2.50
• B. 2.5

Answer 1

Answer: (A), (B)

Horizontal components of velocity are
$u_{1x}=u_1\cos \theta =10\times \cos {37}^{\circ }=8\text{m/s}$
$u_{2x}=u_2\cos \theta =10\times \cos {53}^{\circ }=6\text{m/s}$
and
Vertical component of velocity are
$u_{1y}=u_1\sin \theta =10\sin {37}^{\circ }=6\text{m/s}$
$u_{2y}=u_2\sin \theta =10\sin {53}^{\circ }=8\text{m/s}$

Let mass of each particle be $m$.
Then,
Vertical component of velocity of $COM$ is
$\left(v_{ycom}\right)=\frac{u_{1y}{m}_1+u_{2y}{m}_2}{m_1+m_2}$
$=\frac{6m+8m}{2m}=7\text{m/s}$

$\therefore$ Maximum height attained by COM is
$H_{com}=\frac{(v_{ycom}{)}^2}{2g}=2.5\text{m}$