# The water drops fall at regular intervals from a tap 5 m above the ground. The third drop is leaving the tap at an instant when the first drop touches the ground. How far above the ground is the second drop at that instant? (Take g = 10 m/s-1)            (a)    1.25 m         (b)    2.50 m         (c)    3.75 m         (d)    5.00 m

Amit Saxena
35 Points
10 years ago

(c)

Height of tap = 5m and (g) = 10 m/sec2.

For the first drop, 5 = ut +  gt 2

= (0 × t) +  × 10 t2 v= 5t2 or t2 = 1 or t = 1.

It means that the third drop leaves after one second of the first drop. Or, each drop leaves after every 0.5 sec.

Distance covered by the second drop in 0.5 sec

= (0.5)2 = 1.25m

Therefore, distance of the second drop above the ground

= 5 - 1.25 = 3.75 m.

Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

Height of tap = 5m and (g) = 10 m/sec2.
For the first drop, 5 = ut + gt 2
= (0 × t) + 0.5*10*t^2
= 5t^2
t^2 = 1
t = 1.
It means that the third drop leaves after one second of the first drop.
Or, each drop leaves after every 0.5 sec.
Distance covered by the second drop in 0.5 sec
= (0.5)^2
= 1.25m
Therefore, distance of the second drop above the ground
= 5 - 1.25
= 3.75 m.

Thanks and Regards