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(B) Let the first ball be dropped from rest from height $h$ at $t=0$ and the second ball be dropped from the same height $h$ at $t=1 \,s$.The distance

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(B) Let the first ball be dropped from rest from height $h$ at $t=0$ and the second ball be dropped from the same height $h$ at $t=1 \,s$.The distance covered by the first ball at time $t$ is $H_1 = \frac{1}{2} g t^2$.The second ball is dropped at $t=1 \,s$. Let the time elapsed after the second ball is dropped be $t_1$. Then $t = 1 + t_1$.The distance covered by the first ball at time $t$ is $s_1 = \frac{1}{2} g t^2 = \frac{1}{2} g (1 + t_1)^2$.The distance covered by the second ball at time $t_1$ is $s_2 = \frac{1}{2} g t_1^2$.The distance between the two balls is given by $s_1 - s_2 = 10 \,m$.Substituting the expressions: $\frac{1}{2} g (1 + t_1)^2 - \frac{1}{2} g t_1^2 = 10$.Using $g = 10 \,m/s^2$: $5(1 + 2t_1 + t_1^2) - 5t_1^2 = 10$.$5 + 10t_1 + 5t_1^2 - 5t_1^2 = 10$.$10t_1 = 5$.$t_1 = 0.5 \,s$.The total time $t = 1 + t_1 = 1 + 0.5 = 1.5 \,s$.

$A$ ball is dropped from rest at time $t=0$ from a certain height. $A$ second ball is dropped from the same height at time $t=1 \,s$. At what time $t$, does the distance between the two balls become $10 \,m$ (in $s$)?

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