Two particles of same mass are projected simultaneously with same speed 20 ms^–1 from the top of a tower of height 20m. One is projected vertically upwards and other projected horizontally. The maximum height attained by centre of mass from the ground will be (g = 10 ms^–2)

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10 $\sqrt{2}$ m

25 m

25 $\sqrt{2}$ m

5 m

answer is B.

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Detailed Solution

${\bar{v}}_{cm} = \frac{m_{1} {\bar{v}}_{1} + m_{2} {\bar{v}}_{2}}{m_{1} + m_{2}}$

$= \frac{m (20 \hat{j}) + m (20 \hat{i})}{2 m}$

$v_{cm} = \sqrt{{(10)}^{2} + {(10)}^{2}} = 10 \sqrt{2} m / s$

$tanθ = \frac{v_{1}}{v_{2}} = 1 \Rightarrow θ = 45^{o}$

$H_{\max} = \frac{{v^{2}}_{cm} \sin^{2} θ}{2 g} = \frac{{(10 \sqrt{2})}^{2} x \sin^{2} (45^{o})}{2 x 10} = 5 m$
from ground CM distance = h + H_max = 20 + 5 = 25m

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