Two Particles of same mass are projected simultaneously with same speed $20 m s^{- 1}$ from the top of a tower of height 20m. One of projected vertically upwards and the other projected horizontally. The maximum height(in meters) attained by centre of mass from the ground will be (g=10ms^-2)

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answer is 25.

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Detailed Solution

${\bar{V}}_{CM} = \frac{\sqrt{(mv)^{2} + (mv)^{2}}}{2 m} = \frac{20}{\sqrt{2}}$

$H_{max} = \frac{u^{2} \sin^{2} θ}{2 g} = 5 m$

Max height attained by CM=20+5=25m

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