Question 49: Motion in 1D

Question

A ball is thrown up with a certain velocity so that it reaches a height 'h'. Find the ratio of the two different times of the ball reaching $${h \over 3}$$ in both the directions.

Accepted Answer

$$u = \sqrt {2gh} $$
Now,
$$S = {h \over 3}$$
a = $$-$$g
$$S = ut + {1 \over 2}a{t^2}$$
$${h \over 3} = \sqrt {2gh} t + {1 \over 2}( - g){t^2}$$
$${t^2}\left( {{g \over 2}} \right) - \sqrt {2gh} t + {h \over 3} = 0$$
From quadratic equation
$${t_1},{t_2} = {{\sqrt {2gh}  \pm \sqrt {2gh - {{4g} \over 2}{h \over 3}} } \over g}$$
$${{{t_1}} \over {{t_2}}} = {{\sqrt {2gh}  - \sqrt {{{4gh} \over 3}} } \over {\sqrt {2gh}  + \sqrt {{{4gh} \over 3}} }} = {{\sqrt 3  - \sqrt 2 } \over {\sqrt 3  + \sqrt 2 }}$$

Answer

$${{\sqrt 2  - 1} \over {\sqrt 2  + 1}}$$

Answer

$${1 \over 3}$$

Answer

$${{\sqrt 3  - \sqrt 2 } \over {\sqrt 3  + \sqrt 2 }}$$

Answer

$${{\sqrt 3  - 1} \over {\sqrt 3  + 1}}$$

Motion in 1D

A ball is thrown up with a certain velocity so that it reaches a height 'h'. Find the ratio of the two different times of the ball reaching ${h \over 3}$ in both the directions.

Choose the correct answer: