($g = 10 \text{ m/s}^2$)
This problem involves calculating the position of a falling water drop based on regular time intervals between drops and the total distance to the floor.
Falling Time Determination
Let $h$ be the distance to the floor ($h=5$ m) and $g$ be the acceleration due to gravity ($g=10 \text{ m/s}^2$). The time of flight ($t_f$) for the first drop to reach the floor is found using the equation of motion:
$h = \frac{1}{2} g t_f^2$
Substituting the values:
$5 = \frac{1}{2} (10) t_f^2$
$5 = 5 t_f^2$
$t_f^2 = 1$
$t_f = 1 \text{ s}$
So, the first drop strikes the floor after 1 second.
Time Interval Calculation
The problem states that the first drop strikes the floor exactly when the sixth drop begins to fall. This means the time of flight of the first drop ($t_f$) is equal to the time elapsed from the release of the first drop until the release of the sixth drop. If the time interval between consecutive drops is $T$, then the sixth drop is released at time $5T$ after the first drop (drops are released at $t=0, T, 2T, 3T, 4T, 5T$).
Therefore, $t_f = 5T$.
Using $t_f = 1 \text{ s}$:
$1 = 5T$
$T = \frac{1}{5} = 0.2 \text{ s}$
The time interval between any two consecutive drops is $0.2$ seconds.
Fourth Drop's Fall Time
We need to find the height of the fourth drop when the first drop strikes the ground (at $t = t_f = 1 \text{ s}$).
The drops are released at the following times:
- 1st drop: $t=0$ s
- 2nd drop: $t=T=0.2$ s
- 3rd drop: $t=2T=0.4$ s
- 4th drop: $t=3T=0.6$ s
At the instant the first drop strikes the floor ($t = 1 \text{ s}$), the fourth drop has been falling for a duration ($\Delta t$):
$\Delta t = t_f - (\text{release time of 4th drop})$
$\Delta t = 1 \text{ s} - 0.6 \text{ s} = 0.4 \text{ s}$
Fourth Drop's Fallen Distance
Now, we calculate the distance ($h_4$) fallen by the fourth drop in $0.4$ seconds using the equation of motion:
$h_4 = \frac{1}{2} g (\Delta t)^2$
$h_4 = \frac{1}{2} (10) (0.4)^2$
$h_4 = 5 \times (0.16)$
$h_4 = 0.8 \text{ m}$
Height from Ground
The height of the fourth drop from the ground ($H_4$) is the total height minus the distance it has fallen:
$H_4 = h - h_4$
$H_4 = 5 \text{ m} - 0.8 \text{ m}$
$H_4 = 4.2 \text{ m}$
Thus, the height of the fourth drop from the ground is $4.2$ m.
Similar Questions
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Choose the correct answer from the options given below:
Important Questions from Mechanics
- When both jaws of vernier callipers touch each other, zero mark of the vernier scale is right to zero mark of main scale, $4^{\text{th}}$ mark on vernier scale coincides with certain mark on the main scale. While measuring the length of a cylinder, observer observes $15$ divisions on main scale and $5^{\text{th}}$ division of vernier scale coincides with a main scale division. Measured length of cylinder is ________mm.
(Least count of Vernier calliper = $0.1 \text{ mm}$) - Two wires $A$ and $B$ made of different materials of lengths $6.0 \text{ cm}$ and $5.4 \text{ cm}$, respectively and area of cross sections $3.0 \times 10^{-5} \text{ m}^2$ and $4.5 \times 10^{-5} \text{ m}^2$, respectively are stretched by the same magnitude under a given load. The ratio of the Young's modulus of $A$ to that of $B$ is $x : 3$. The value of $x$ is________.
Two circular discs of radius each $10 \text{ cm}$ are joined at their centres by a rod of length $30 \text{ cm}$ and mass $600 \text{ gm}$ as shown in figure.
If the mass of each disc is $600 \text{ gm}$ and applied torque between two discs is $43 \times 10^5 \text{ dyne.cm}$, the angular acceleration of the discs about the given axis $AB$ is________$\text{rad/s}^2$.- A particle of mass $m$ falls from rest through a resistive medium having resistive force, $F = -kv$, where $v$ is the velocity of the particle and $k$ is a constant. Which of the following graphs represents velocity ($v$) versus time ($t$)?
- A block of mass $5 \text{ kg}$ is moving on an inclined plane which makes an angle of $30^\circ$ with the horizontal. Friction coefficient between the block and inclined plane surface is $\frac{\sqrt{3}}{2}$. The force to be applied on the block so that the block will move down without acceleration is ________$\text{N}$.
($g = 10 \text{ m/s}^2$).
