Question
A person sitting on the top of a tall building is dropping balls at regular intervals of one second. Find the positions of the $3^{rd}, 4^{th}$ and $5^{th}$ ball when the $6^{th}$ ball is being dropped.
✓
Answer
For every ball, $u = 0, a = g = 9.8m/s^2$
$\therefore 4^{th}$ ball move for $2 \sec$, $5^{th}$ ball $1 \sec$ and $3^{rd}$ ball $3 \sec$ when $6^{th}$ ball is being dropped.
For $3^{rd}$ ball t = 3 sec$\text{S}_3=\text{ut}+\frac{1}{2}\text{at}^2=0+\frac{1}{2}(9.8)3^2=4.9\text{m}$ below the top.
For $4^{th}$ ball, t = 2 sec$\text{S}_2=0+\frac{1}{2}\text{ gt}^2=\frac{1}{2}(9.8)2^2=19.6\text{m}$ below the top (u = 0)
For $5^{th}$ ball, t = 1 sec$\text{S}_3=\text{ut}+\frac{1}{2}\text{at}^2=0+\frac{1}{2}(9.8)\text{t}^2=4.98\text{m}$ below the top.
$\therefore 4^{th}$ ball move for $2 \sec$, $5^{th}$ ball $1 \sec$ and $3^{rd}$ ball $3 \sec$ when $6^{th}$ ball is being dropped.
For $3^{rd}$ ball t = 3 sec$\text{S}_3=\text{ut}+\frac{1}{2}\text{at}^2=0+\frac{1}{2}(9.8)3^2=4.9\text{m}$ below the top.
For $4^{th}$ ball, t = 2 sec$\text{S}_2=0+\frac{1}{2}\text{ gt}^2=\frac{1}{2}(9.8)2^2=19.6\text{m}$ below the top (u = 0)
For $5^{th}$ ball, t = 1 sec$\text{S}_3=\text{ut}+\frac{1}{2}\text{at}^2=0+\frac{1}{2}(9.8)\text{t}^2=4.98\text{m}$ below the top.
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