A stone is thrown up from the mouth of a well at g m/s. as it passes the mouth into the well another stone is dropped alongside with first stone. Two splashes are heard at an interval of 1 second . the depth of the well is ___

To solve the problem of determining the depth of the well, we need to analyze the motion of both stones: the one thrown upwards and the one dropped downwards. The key to this problem lies in understanding the time it takes for each stone to reach the water and the time interval between the two splashes.

Breaking Down the Problem

Let's denote the following variables:

• g: acceleration due to gravity (in m/s²)
• h: depth of the well (in meters)
• t_1: time taken by the first stone to reach the water (in seconds)
• t_2: time taken by the second stone to reach the water (in seconds)
• t: time interval between the two splashes (1 second)

Analyzing the First Stone

The first stone is thrown upwards with an initial velocity. As it ascends, it will eventually stop and then fall back down. The time taken to reach the highest point can be calculated using the formula:

t_{up} = \frac{v}{g}

where v is the initial velocity of the stone. After reaching the highest point, the stone will fall back down to the water. The total time t_1 for the first stone to reach the water can be expressed as:

t_1 = t_{up} + t_{down}

The time taken to fall back down from the highest point to the water can be calculated using the equation of motion:

h = \frac{1}{2} g t_{down}^2

Analyzing the Second Stone

The second stone is simply dropped, so its motion is straightforward. The time taken for the second stone to reach the water is given by:

h = \frac{1}{2} g t_2^2

Setting Up the Equations

Since we know that the interval between the two splashes is 1 second, we can express this relationship as:

t_1 - t_2 = 1

Now, substituting the expressions for t_1 and t_2 into this equation will help us find the depth of the well.

Calculating the Depth

From the first stone, we have:

t_1 = \frac{v}{g} + \sqrt{\frac{2h}{g}}

From the second stone:

t_2 = \sqrt{\frac{2h}{g}}

Substituting t_2 into the interval equation gives:

\frac{v}{g} + \sqrt{\frac{2h}{g}} - \sqrt{\frac{2h}{g}} = 1

This simplifies to:

\frac{v}{g} = 1

From this, we can express the initial velocity:

v = g

Now, substituting this back into the equation for t_1:

t_1 = \frac{g}{g} + \sqrt{\frac{2h}{g}} = 1 + \sqrt{\frac{2h}{g}}

Setting this equal to t_2 + 1 gives:

1 + \sqrt{\frac{2h}{g}} = \sqrt{\frac{2h}{g}} + 1

From this, we can derive the depth of the well:

h = \frac{g}{2}

Final Thoughts

Thus, the depth of the well can be expressed in terms of the acceleration due to gravity. If we substitute the value of g (approximately 9.81 m/s²), we can find the numerical value of h. Therefore, the depth of the well is:

h = \frac{9.81}{2} \approx 4.905 \text{ meters}

This approach illustrates how we can analyze the motion of objects under gravity and use their interactions to derive useful information about their environment. If you have any further questions or need clarification on any steps, feel free to ask!