Question

A particle projected vertically upwards attains a maximum height H. If the ratio of the times to attain a height $ h(h

• A. $ 4h=3H $
• B. $ 3h=4H $
• C. $ 3h=H $
• D. $ 4h=H $

Answer 1

Answer: (A)

From the equation of motion $ {{v}^{2}}={{u}^{2}}+2as $ $ {{v}^{2}}-{{u}^{2}}=2as $ $ {{0}^{2}}-{{u}^{2}}=2\times (-g)\times H $ $ u=\sqrt{2gH} $ Now, $ h=ut-\frac{1}{2}g{{t}^{2}} $ $ h=(\sqrt{2gH})t-\frac{1}{2}g{{t}^{2}} $ This equation is quadratic in t. Solve for t we get, values for time. Ratio is $ \frac{{{t}_{1}}}{{{t}_{2}}}=\frac{1}{3} $ (given) substituting values, we get $ 4h=3H $