COMEDK 2021 | Motion Question 13 | Physics

The displacement $$x$$ of a particle varies with time $$t$$, $$x=ae^{-pt}+be^{qt}$$, where a, b, p and q are positive constant. The velocity of the particle will

go on increasing forever

be independent of p and q

drop to zero when p = q

go on decreasing with time

COMEDK 2021

MCQ (Single Correct Answer)

-0

The angle of projection with the horizontal in terms of maximum height attained and horizontal range is given by

$${\tan ^{ - 1}}\left( {{{2H} \over {3R}}} \right)$$

$${\tan ^{ - 1}}\left( {{{4R} \over {3H}}} \right)$$

$${\tan ^{ - 1}}\left( {{{4H} \over R}} \right)$$

$${\tan ^{ - 1}}\left( {{R \over H}} \right)$$

COMEDK 2020

MCQ (Single Correct Answer)

-0

A body is projected vertically upwards. The times corresponding to height $$h$$, while ascending and while descending are $$t_1$$ and $$t _2$$ , respectively. Then, the velocity of projection is ($$g$$ is acceleration due to gravity)

$${{g\sqrt {{t_1}{t_2}} } \over 2}$$

$${{g({t_1} + {t_2})} \over 2}$$

$$g\sqrt {{t_1}{t_2}} $$

$${{g{t_1}{t_2}} \over {{t_1} + {t_2}}}$$

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