Question

Water drops fall at regular intervals from a tap 5 m above the ground. How far above the ground is the second drop at that instant when the 3rd drop left the tap and $1_{st}$ drop reaches the ground. (g= $10m/s^2$)

• A. 1.25 m
• B. 2.50 m
• C. 3.75 m
• D. 4.00 m

Answer 1

The correct option is C 3.75 m

The time after which 1st drop leaves tap and reaches ground is,

$t_1=\sqrt{\frac{2h}{g}}=\sqrt{\frac{10}{10}}=1s$

At this instant the 3rd drop has just left the tap. This means the instant at which 2nd drop left the tap is,

$t_2=t_1/2=0.5s$ earlier.

Initial velocity of the drop $u=0$

Thus from $S=u{t}_2+\frac{1}{2}g{t}_2^2$

So distance covered by 2nd drop from tap is, $S=0.5g{t}_2^2=\left(0.5\right)\left(10\right)(0.5{)}^2=1.25m$

So height of second drop above ground is, $h=5-1.25=3.75m$