wo stones are dropped from top of tower a half a second apart.the time after dropping the first stone at which the distance between two stones is 20m is

To solve the problem of two stones dropped from the top of a tower, we need to analyze their motion under the influence of gravity. The first stone is dropped at time \( t = 0 \) seconds, and the second stone is dropped half a second later, at \( t = 0.5 \) seconds. We want to find the time after dropping the first stone when the distance between the two stones is 20 meters.

Understanding the Motion of the Stones

Both stones are in free fall, which means they will accelerate downwards at a rate of approximately \( 9.81 \, \text{m/s}^2 \) (the acceleration due to gravity). The distance fallen by an object in free fall can be calculated using the formula:

d = \frac{1}{2} g t^2

where:

• d is the distance fallen in meters
• g is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
• t is the time in seconds since the object was dropped

Calculating the Distances

Let's denote the time after the first stone is dropped as \( t \). The distance fallen by the first stone after time \( t \) is:

d_1 = \frac{1}{2} g t^2

For the second stone, which is dropped half a second later, the time it has been falling when the first stone has been falling for \( t \) seconds is \( t - 0.5 \) seconds. Therefore, the distance fallen by the second stone is:

d_2 = \frac{1}{2} g (t - 0.5)^2

Finding the Distance Between the Two Stones

The distance between the two stones at any time \( t \) can be expressed as:

Distance = d_1 - d_2

Substituting the expressions for \( d_1 \) and \( d_2 \), we have:

Distance = \frac{1}{2} g t^2 - \frac{1}{2} g (t - 0.5)^2

We want this distance to equal 20 meters:

\frac{1}{2} g t^2 - \frac{1}{2} g (t - 0.5)^2 = 20

Solving the Equation

First, we can factor out \( \frac{1}{2} g \):

\frac{1}{2} g (t^2 - (t - 0.5)^2) = 20

Next, we need to simplify \( (t - 0.5)^2 \):

(t - 0.5)^2 = t^2 - t + 0.25

Substituting this back into the equation gives:

\frac{1}{2} g (t^2 - (t^2 - t + 0.25)) = 20

This simplifies to:

\frac{1}{2} g (t + 0.25) = 20

Now, substituting \( g \approx 9.81 \, \text{m/s}^2 \):

\frac{1}{2} \times 9.81 \times (t + 0.25) = 20

Multiplying both sides by 2 gives:

9.81(t + 0.25) = 40

Now, divide both sides by 9.81:

t + 0.25 = \frac{40}{9.81}

Calculating \( \frac{40}{9.81} \) gives approximately 4.08:

t + 0.25 \approx 4.08

Subtracting 0.25 from both sides results in:

t \approx 3.83 \, \text{seconds}

Final Result

Thus, the time after dropping the first stone at which the distance between the two stones is 20 meters is approximately 3.83 seconds. This means that after about 3.83 seconds from the moment the first stone is dropped, the second stone, which was dropped half a second later, will be 20 meters away from the first stone.