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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 	imes

Vedclass · Accepted Answer

$95$

Vedclass · Answer

(A) The time taken by the first ball to reach the ground is given by $t_1 = \sqrt{\frac{2h}{g}}$.Substituting the values, $t_1 = \sqrt{\frac{2 	imes 500}{10}} = \sqrt{100} = 10 \ s$.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 $s_2 = \frac{1}{2} 	imes g 	imes t_2^2$.$s_2 = \frac{1}{2} 	imes 10 	imes 9^2 = 5 	imes 81 = 405 \ m$.The distance between the two balls is $s_1 - s_2 = 500 - 405 = 95 \ m$.

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 $g = 10 \ m/s^2$)

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