Four marbles are dropped from the top of a to | vedclass

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(A) Let the acceleration due to gravity be $g = 10 \, m/s^2$.The first marble is dropped at $t = 0$ and reaches the ground at $t = 4 \, s$. Its distan

Vedclass · Accepted Answer

$35, 25$ and $15 \, m$

Vedclass · Answer

(A) Let the acceleration due to gravity be $g = 10 \, m/s^2$.The first marble is dropped at $t = 0$ and reaches the ground at $t = 4 \, s$. Its distance from the top is $h_1 = \frac{1}{2} g (4)^2 = 8g = 80 \, m$.The second marble is dropped at $t = 1 \, s$,so at $t = 4 \, s$,it has been falling for $3 \, s$. Its distance is $h_2 = \frac{1}{2} g (3)^2 = 4.5g = 45 \, m$.The third marble is dropped at $t = 2 \, s$,so at $t = 4 \, s$,it has been falling for $2 \, s$. Its distance is $h_3 = \frac{1}{2} g (2)^2 = 2g = 20 \, m$.The fourth marble is dropped at $t = 3 \, s$,so at $t = 4 \, s$,it has been falling for $1 \, s$. Its distance is $h_4 = \frac{1}{2} g (1)^2 = 0.5g = 5 \, m$.The distance between the first and second marble is $h_1 - h_2 = 80 - 45 = 35 \, m$.The distance between the second and third marble is $h_2 - h_3 = 45 - 20 = 25 \, m$.The distance between the third and fourth marble is $h_3 - h_4 = 20 - 5 = 15 \, m$.Thus,the distances are $35 \, m, 25 \, m$ and $15 \, m$.

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