A ball is thrown vertically upward with such speed that it reaches maximum height h. Find the ratio of the times at which it is at height h/3 while going up and coming down, respectively.
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Question

A ball is thrown vertically upward with such speed that it reaches maximum height h. Find the ratio of the times at which it is at height h/3 while going up and coming down, respectively.

G1 Physics · Accepted Answer

Maximum height h=u²/(2g). At y=h/3, solve ut-gt²/2=h/3. The two roots are t=(u/g)[1±√(2/3)]. Hence ascent time/descent time=[1-√(2/3)]/[1+√(2/3)]=(√3-√2)/(√3+√2). Hence option B.