Question

Water drops fall at regular intervals from a tap, which is 5 m above the ground. The third drop is leaving the tap at the instant the first drop reaches the ground. At this instant, the height(in m) of the second drop from the ground is

Answer 1

Answer:

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

2

For the first drop, 5=ut+gt

2

=(0×t)+×10t

2v

=5t

2

or t

2

=1 or t=1.

It mean 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=3.75m

Answer 2

Answer:

Explanation:

simply calculate time it reaches ground

s=1/2at^2

(2s/a)^1/2 =t

t=1

as drops in regular intervals of  time

time drops are fallen after over after = 1/2

displacement of  drop 2 from point of projection = 10/8 m

displacement of drop 2 from ground = 30/8 = 3.75