$A$ ball of mass $0.5 \; kg$ is dropped from a height of $10 \; m$. The height from the ground,at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity,is $\dots \; m$. (Use $g = 10 \; m/s^2$).
- A$1$
- B$3$
- C$5$
- D$7$
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Similar Questions
$A$ balloon rises up with a constant net acceleration of $10\,m/s^2$. After $2\,s$,a particle drops from the balloon. After a further $2\,s$,match the following columns. (Take $g=10\,m/s^2$)
Column $I$ Column $II$ $(A)$ Height of particle from ground $(p)$ $0$ $(B)$ Speed of particle $(q)$ $10$ $SI$ unit $(C)$ Displacement of particle $(r)$ $40$ $SI$ unit $(D)$ Acceleration of particle $(s)$ $20$ $SI$ unit
| Column $I$ | Column $II$ |
|---|---|
| $(A)$ Height of particle from ground | $(p)$ $0$ |
| $(B)$ Speed of particle | $(q)$ $10$ $SI$ unit |
| $(C)$ Displacement of particle | $(r)$ $40$ $SI$ unit |
| $(D)$ Acceleration of particle | $(s)$ $20$ $SI$ unit |
Difficult
View Solution$A$ body falls freely for $10 \ s$. What is its average velocity during this journey (in $m/s$)? (Take $g = 10 \ m/s^2$)
$A$ balloon is at a height of $81\, m$ and is ascending upwards with a velocity of $12\, m/s$. $A$ body of $2\, kg$ weight is dropped from it. If $g = 10\, m/s^2$,the body will reach the surface of the earth in ......... $s$.
Medium
View SolutionThe time of flight of a vertically projected stone is $8 \,s$. The position of the stone after $6 \,s$ from the ground is (Acceleration due to gravity $= 10 \,m/s^2$). (in $\,m$)
$A$ body dropped from the top of a tower clears $7/16$ of the total height of the tower in its last second of flight. The time taken by the body to reach the ground is........$s$
Medium
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