A stone of mass `5kg` falls from the top of a cliff `50m` high and buries `1m` deep in sand . Find the average resistance offered by the sand and the time it takes to penetrate .

Explanation

To solve the problem, we need to find two things: the average resistance offered by the sand and the time it takes for the stone to penetrate 1 meter deep into the sand after falling from a height of 50 meters.

### Step 1: Calculate the velocity of the stone just before it hits the sand.

We can use the equation of motion for free fall:
\[ v^2 = u^2 + 2gh \]
where:
- \( v \) = final velocity (just before hitting the sand)
- \( u \) = initial velocity (0 m/s, since it starts from rest)
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( h \) = height (50 m)

Substituting the values:
\[ v^2 = 0 + 2 \times 9.81 \times 50 \]
\[ v^2 = 981 \]
\[ v = \sqrt{981} \approx 31.32 \, \text{m/s} \]

### Step 2: Calculate the deceleration of the stone while penetrating the sand.

When the stone penetrates the sand, it comes to a stop after traveling 1 meter. We can use the equation of motion again:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) = final velocity (0 m/s, since it comes to rest)
- \( u \) = initial velocity (31.32 m/s, just before hitting the sand)
- \( a \) = acceleration (deceleration in this case)
- \( s \) = distance penetrated (1 m)

Rearranging the equation to solve for \( a \):
\[ 0 = (31.32)^2 + 2a(1) \]
\[ 0 = 981 + 2a \]
\[ 2a = -981 \]
\[ a = -490.5 \, \text{m/s}^2 \]

### Step 3: Calculate the average resistance offered by the sand.

The average resistance \( F_r \) can be calculated using Newton's second law:
\[ F_r = m \cdot a \]
where:
- \( m \) = mass of the stone (5 kg)
- \( a \) = deceleration (490.5 m/s²)

Substituting the values:
\[ F_r = 5 \times 490.5 \]
\[ F_r = 2452.5 \, \text{N} \]

### Step 4: Calculate the time taken to penetrate the sand.

We can use the equation of motion:
\[ v = u + at \]
Rearranging to solve for time \( t \):
\[ t = \frac{v - u}{a} \]
Substituting the values:
\[ t = \frac{0 - 31.32}{-490.5} \]
\[ t = \frac{-31.32}{-490.5} \]
\[ t \approx 0.064 \, \text{s} \]

### Final Results:
- Average resistance offered by the sand: **2452.5 N**
- Time taken to penetrate the sand: **0.064 seconds**

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