A person sitting on the top of a tall building is dropping balls at regular intervals of one second. When the $6^{t h}$ ball is being dropped, the positions of the $3^{r d}$ , $4^{t h}$ and $5^{t h}$ balls from the top of the building are respectively:

4.9 m, 19.6 m, 44.1 m

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4.9 m, 14.7 m, 24.5 m

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44.1 m, 19.6 m, 4.9 m

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24.5 m, 14.7 m, 4.9 m

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Solution

The correct option is C 44.1 m, 19.6 m, 4.9 m
Position of $3^{r d}$ ball:
$s_{3} = u t + \frac{1}{2} g t^{2} = \frac{1}{2} g (3)^{2} = \frac{9 g}{2} = 44.1$
$s_{4} = \frac{1}{2} g (2)^{2} = 2 g = 19.6$
$s_{5} = \frac{1}{2} g (1)^{2} = \frac{g}{2} = 4.9$

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A person sitting on the top of a tall building is dropping balls at regular intervals of one second. When the 6th ball is being dropped, the positions of the 3rd, 4th and 5th balls from the top of the building are respectively:

The correct option is C 44.1 m, 19.6 m, 4.9 mPosition of 3rd ball:s3=ut+12gt2=12g(3)2=9g2=44.1s4=12g(2)2=2g=19.6s5=12g(1)2=g2=4.9

A person sitting on the top of a tall building is dropping balls at regular intervals of one second. When the $6^{t h}$ ball is being dropped, the positions of the $3^{r d}$ , $4^{t h}$ and $5^{t h}$ balls from the top of the building are respectively:

The correct option is C 44.1 m, 19.6 m, 4.9 m
Position of $3^{r d}$ ball:
$s_{3} = u t + \frac{1}{2} g t^{2} = \frac{1}{2} g (3)^{2} = \frac{9 g}{2} = 44.1$
$s_{4} = \frac{1}{2} g (2)^{2} = 2 g = 19.6$
$s_{5} = \frac{1}{2} g (1)^{2} = \frac{g}{2} = 4.9$