Question

Q5
Water drops from a leaky tap are falling, at regular time intervals, on the ground 125 cm
below the tap. The first drop strikes the ground when the sixth drop just begins to fall. What
is the height from the ground of the third drop when the first drop strikes the ground? Take
g = 10 ms 2
50 cm
60 cm
70 cm
80 cm

Answer 1

Answer:

answer is 50 cm option A is correct

Answer 2

Concept:

• One-dimensional motion
• Kinematics equations

Given:

• Height of the tap above the ground = 125cm = 1.25 m
• Initial velocity of drops = 0 m/s
• When the sixth drop starts to fall, the first drop reaches the ground

Find:

• The height from the ground of the third drop when the first one reaches the ground

Solution:

s = ut+1/2at^2

Consider the first drop

1.25 = 0+1/2(10)t^2

1.25 = 5t^2

0.25 = t^2

t = 0.5s

It takes 0.5 s for the first drop to reach the ground. Thus, the sixth drop leaves 0.5s after the first drop. So, in 0.5s, 5 drops have fallen. Each drop leaves after 0.1s.

The 3rd drop leaves after 0.2s and the first drop strikes the ground at 0.5s, so when the first drop strikes the ground, the 3rd drop covers a certain distance in the time 0.5-0.2 = 0.3s

s = ut+1/2gt^2

s = 0+1/2(10)(0.09)

s = 0.45 m

The height from the ground is 1.25-0.45 = 0.8m

The height from the ground of the third drop when the first one reaches the ground is 0.8m

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