Q. Consider a tap situated on the top of a $5m$ high tower. What is the position of the $2^{nd}$ drop if it is known that, the $3^{rd}$ drop leaves the tap when the $1^{st}$ drop reaches the ground?

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A

$1.25m$

B

$2.50m$

C

$3.75m$

D

$4.00m$

Solution:

$t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 5}{10}}=1s$

Let t₀ is the interval between two drops. then
2 t₀ = t
t₀ = 0.5 s
2nd drop has taken t0 time to fall. Therefore distance fallen,
$d=\frac{1}{2}gt_{0}^{2}=\left(\frac{1}{2}\right)\left(\right.10\left.\right)\left(\right.0.5\left(\left.\right)^{2}$
= 1.25 m
Height from ground = h - d
= 5 - 1.25
= 3.75 m

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