Two particles of the same mass are projected from the ground with the same speed 10 m - s at an angle 37- and 53- with the horizontal respectively- Find the maximum height attained by their COM in metres- Take g -9-8 m - s 2 and answer upto two decimal places-

Horizontal components of velocity are nbsp; u _1 x = u_1cosθ = 10 ×cos37^∘ =8 m/s u _2 x = u_2cosθ = 10 ×cos53^∘ = 6 m/s and Vertical component of velocity ...

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Byju's Answer

Standard XII

Physics

Velocity of COM

Two particles...

Question

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

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Solution

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

Let mass of each particle be $m$ .
Then,
Vertical component of velocity of $C O M$ is
$(v_{y c o m}) = \frac{u_{1 y} m_{1} + u_{2 y} m_{2}}{m_{1} + m_{2}}$
$= \frac{6 m + 8 m}{2 m} = 7 m/s$

$∴$ Maximum height attained by COM is
$H_{c o m} = \frac{(v_{y c o m})^{2}}{2 g} = 2.5 m$

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Two particles of the same mass are projected from the ground with the same speed 10 m/s at an angle 37∘ and 53∘ with the horizontal respectively. Find the maximum height attained by their COM (in metres). Take (g=9.8 m/s2) and answer upto two decimal places.

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