A ball is released from rest. If it takes 1 second to cross the last 20 m before hitting the ground, find the height from which it was dropped.

Answer

To find the height from which the ball was dropped, we can break down the problem into a series of steps:

### Step 1: Understand the problem
The ball is released from rest, meaning its initial velocity \( u = 0 \). It takes 1 second to cover the last 20 meters before hitting the ground.

### Step 2: Use the equations of motion
We can use the second equation of motion to find the final velocity \( v \) of the ball just before it covers the last 20 meters.

The equation is:
\[ s = ut + \frac{1}{2}gt^2 \]
where:
- \( s = 20 \, \text{m} \) (the distance covered in the last second),
- \( u = 0 \) (initial velocity),
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( t = 1 \, \text{s} \) (time taken to cover the last 20 m).

Substituting the values:
\[ 20 = 0 \cdot 1 + \frac{1}{2} \cdot 10 \cdot (1)^2 \]
\[ 20 = 0 + 5 \]
This confirms that the ball covers 20 m in 1 second.

### Step 3: Calculate the final velocity \( v \)
Using the first equation of motion:
\[ v = u + gt \]
Substituting the known values:
\[ v = 0 + 10 \cdot 1 = 10 \, \text{m/s} \]

### Step 4: Use the third equation of motion to find the height
Now, we can find the distance \( h - 20 \) (the height from which the ball was dropped minus the last 20 m) using:
\[ v^2 = u^2 + 2g s \]
Where:
- \( v = 10 \, \text{m/s} \) (final velocity before the last 20 m),
- \( u = 0 \) (initial velocity),
- \( g = 10 \, \text{m/s}^2 \),
- \( s = h - 20 \) (distance from the top to the point just before the last 20 m).

Substituting the values:
\[ (10)^2 = 0 + 2 \cdot 10 \cdot (h - 20) \]
\[ 100 = 20(h - 20) \]
\[ 100 = 20h - 400 \]
\[ 20h = 500 \]
\[ h = 25 \, \text{m} \]

### Step 5: Total height
Now, add the last 20 m to find the total height from which the ball was dropped:
\[ \text{Total height} = h + 20 = 25 + 20 = 45 \, \text{m} \]

### Final Answer
The height from which the ball was dropped is **45 meters**.

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