Two steel balls fall freely on an elastic slab; the first is dropped from 8 m and the second from 2 m, T seconds later. If the velocities of the balls coincide after the passage of time T , then T equ

Question

Two steel balls fall freely on an elastic slab; the first is dropped from 8 m and the second from 2 m, $T$ seconds later. If the velocities of the balls coincide after the passage of time $T$ , then $T$ equals: (Take $g = 10 m / s^{2}$ )

Answer:

1.27

Solution:

$T = 2 (2 \times 8/10 - 2 \times 2/10) = 2 (1.265 - 0.632) = 1.27 s$ .

TestHub · Accepted Answer

T = 2 ( 2 × 8/10 ​ − 2 × 2/10 ​ ) = 2 ( 1.265 − 0.632 ) = 1.27 s .

Physics - Kinematics Question with Solution | TestHub

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