(A) The time taken by the first ball to reach the ground is given by $t_1 = \sqrt{\frac{2h}{g}}$.Substituting the values, $t_1 = \sqrt{\frac{2 \times

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(A) The time taken by the first ball to reach the ground is given by $t_1 = \sqrt{\frac{2h}{g}}$.Substituting the values, $t_1 = \sqrt{\frac{2 \times 500}{10}} = \sqrt{100} = 10 \ s$.The first ball hits the ground at $t = 10 \ s$.The second ball is released $1 \ s$ later, so it has been in motion for $t_2 = 10 - 1 = 9 \ s$.The distance traveled by the first ball is $s_1 = 500 \ m$ (as it hits the ground).The distance traveled by the second ball in $9 \ s$ is $s_2 = \frac{1}{2} \times g \times t_2^2$.$s_2 = \frac{1}{2} \times 10 \times 9^2 = 5 \times 81 = 405 \ m$.The distance between the two balls is $s_1 - s_2 = 500 - 405 = 95 \ m$.

Class 11 Physics Motion in a Straight Line Motion Under Gravity

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Two balls are released from the same position at a height of $500 \ m$ above the ground, one after the other, with an interval of $1 \ s$. What is the distance between the two balls when the first ball hits the ground (in $m$)? (Acceleration due to gravity $g = 10 \ m/s^2$)

TS EAMCET 2021 All TS EAMCET

A
$95$
B
$65$
C
$130$
D
$175$

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Motion in a Straight Line

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Motion Under Gravity

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$A$ 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

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$A$ ball is thrown vertically upwards with a velocity of $20 \; m s^{-1}$ from the top of a multistorey building. The height of the point from where the ball is thrown is $25.0 \; m$ from the ground. How high will the ball rise above the point of projection (in $; m$)? Take $g = 10 \; m s^{-2}$.

Medium

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$A$ stone falls freely under gravity. It covers distances $h_1, h_2$ and $h_3$ in the first $5 \ s$,the next $5 \ s$ and the next $5 \ s$ respectively. The relation between $h_1, h_2$ and $h_3$ is:

MediumAIPMT 2013

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$A$ body falling under gravity covers two points $A$ and $B$ separated by $80 \ m$ in $2 \ s$. The distance of the upper point $A$ from the starting point is (use $g = 10 \ m/s^2$): (in $m$)

DifficultJEE MAIN 2024

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From a building,two balls $A$ and $B$ are thrown such that $A$ is thrown upwards and $B$ is thrown downwards with the same speed $u$ (both vertically). If $v_{A}$ and $v_{B}$ are their respective velocities on reaching the ground,then:

MediumAIEEE 2002

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Two balls are released from the same position at a height of $500 \ m$ above the ground, one after the other, with an interval of $1 \ s$. What is the distance between the two balls when the first ball hits the ground (in $m$)? (Acceleration due to gravity $g = 10 \ m/s^2$)

Similar Questions

$A$ ball is thrown vertically upwards with a velocity of $20 \; m s^{-1}$ from the top of a multistorey building. The height of the point from where the ball is thrown is $25.0 \; m$ from the ground. How high will the ball rise above the point of projection (in $; m$)? Take $g = 10 \; m s^{-2}$.

$A$ stone falls freely under gravity. It covers distances $h_1, h_2$ and $h_3$ in the first $5 \ s$,the next $5 \ s$ and the next $5 \ s$ respectively. The relation between $h_1, h_2$ and $h_3$ is:

$A$ body falling under gravity covers two points $A$ and $B$ separated by $80 \ m$ in $2 \ s$. The distance of the upper point $A$ from the starting point is (use $g = 10 \ m/s^2$): (in $m$)

From a building,two balls $A$ and $B$ are thrown such that $A$ is thrown upwards and $B$ is thrown downwards with the same speed $u$ (both vertically). If $v_{A}$ and $v_{B}$ are their respective velocities on reaching the ground,then:

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