A ball of mass m is thrown vertically up. Another ball of mass 2m is thrown at an angle with the vertical. Both of them stay in the air for the same duration. Find the ratio of the heights attained by the two balls
Asked by Roopak | 25th Jul, 2013, 10:42: PM
height obtained by an object thrown up with u velocity (h) = u^2/2g
Hence, ratio of the heights (h1/h2) = u1^2/u2^2
Also, since they remain in the air for same duration.
So, using first law of motion v = u + at
v = 0, t = -u/a, Now here, since the a is same as equal to g and t is same, so u needs to be same.
Hence, ratio of heights = 1:1
Answered by | 26th Jul, 2013, 05:28: AM
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