Question

A ball of mass M is thrown vertically of another ball of mass 2 m is thrown at an angle theta with the vertical both of them is stay in air for the time period of same period of time what is the ratio of height attained by two balls

Answer 1

Height (h) obtained with velocity u = u²/2g ratio of the heights (h1/h2) = u1²/u2²

As they remain in the air for same duration. according to first law of motion v = u + atv = 0
t = -u/a
a=g and t is same, so u is also same Hence, ratio of heights = 1:1

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