Question 5: Motion in 1D

Question

Water drops fall from a tap on the floor, 5 m below, at regular intervals of time, the first drop strikes the floor when the sixth drop begins to fall. The height at which the fourth drop will be from ground, at the instant when the first drop strikes the ground is $\_\_\_\_$ m.
$$ \left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right) $$

Accepted Answer

The first drop falls a distance of $H=5 \mathrm{~m}$ starting from rest $(u=0)$.
Using the equation of motion:
$$ \mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^2 $$
$$\Rightarrow $$ $5=0+\frac{1}{2} 	imes 10 	imes \mathrm{T}^2$
$$\Rightarrow $$ $5=5 \mathrm{~T}^2 \Rightarrow \mathrm{~T}^2=1 \Rightarrow \mathrm{~T}=1 \mathrm{~s}$
When the first drop hits the ground, the sixth drop is just beginning to fall. This means there are 5 equal time intervals between the first and sixth drops.
$$ \Delta \mathrm{t}=\frac{\mathrm{T}}{5}=\frac{1}{5}=0.2 \mathrm{~s} $$
At the instant the first drop strikes the ground $(\mathrm{t}=1 \mathrm{~s})$ :
Drop 1 has fallen for 1.0 s
Drop 2 has fallen for 0.8 s
Drop 3 has fallen for 0.6 s
Drop 4 has fallen for 0.4 s
Using the distance formula for the fourth drop:
$$ \mathrm{h}_4=\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_4ight)^2 $$
$$\Rightarrow $$ $h_4=\frac{1}{2} 	imes 10 	imes(0.4)^2$
$$\Rightarrow $$ $$ \mathrm{h}_4=5 	imes 0.16=0.8 \mathrm{~m} $$
The height from the ground is the total height minus the distance fallen:
Height from ground $=\mathrm{H}-\mathrm{h}_4=5 \mathrm{~m}-0.8 \mathrm{~m}=4.2 \mathrm{~m}$
Hence, the correct option is (C).

Answer

3.8

Answer

4.0

Answer

4.2

Answer

2.5

Motion in 1D

Choose the correct answer: