A person siting on the top of a ball building is dropping balls at regular intervals of one second. Find the positions of the 3rd, 4th and 5th ball when the 6th ball is being dropped.

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Solution

For every ball, u = 0, a = g = 9.8 $m / s^{2}$ 4th ball moves for 2 sec. 5th ball for 1 sec and 3rd ball for 3 sec when 6th ball is being dropped.

For 3rd ball, t = 3 sec

$∴ S_{1} = u t + \frac{1}{2 a t^{2}}$

= $0 + \frac{1}{2} 9.8 (3)^{2}$

= 44.1 m below the top

For 4th ball, t = 2 sec

$∴ S_{2} = 0 + \frac{1}{2} g t^{2} = \frac{1}{2} \times 9.8 \times (2)^{2}$

= 19.6 m below the top.

[u = 0]]

For 5th ball, t = 1 sec

$∴ S_{3} = u t + \frac{1}{2} a t^{2}$

= $0 + \frac{1}{2} \times 9.8 \times t^{2}$

= 4.9 m below the top

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