KinematicsMedium
Question
Water drops fall from a tap on the floor, 5 m below, at regular intervals of time, the first drop strikes the floor when the sixth drop begins to fall. The height at which the fourth drop will be from ground, at the instant when the first drop strikes the ground is $\_\_\_\_$ m.
$$\left( g = 10\text{ }m/s^{2} \right) $$
Options
A.2.5
B.4.0
C.4.2
D.3.8
Solution
Time to reach ground $= \sqrt{\frac{2\text{ }h}{\text{ }g}} = \sqrt{\frac{2 \times 5}{10}} = 1sec$
Five drops per second
Time between each drop $= 0.2sec$.
Time of fall for $4^{\text{th~}}$ drop is $1 - 0.6 = 0.4sec$
Height fall of $4^{\text{th~}}$ drop is $= \frac{1}{2} \times 10 \times {0.4}^{2} = 0.8\text{ }m$
Height from ground $= 5 - 0.8 = 4.2\text{ }m$
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