Question

A particle projected vertically upwards attains a maximum height H. If the ratio of the times to attain a height $h(h<H)$ is $\frac{1}{3}$ then,

• 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 $4h=3H$