A ball is dropped from a height. If it takes 0.200 s to cross thelast 6.00 m before hitting the ground, find the height from which it was dropped. `Take g=10 m/s^2`.

Explanation

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

### Step 1: Understand the Problem
We know that a ball is dropped from a height and takes 0.200 seconds to cross the last 6.00 meters before hitting the ground. We need to find the total height from which it was dropped.

### Step 2: Use Kinematic Equations
We will use the second equation of motion to relate displacement, initial velocity, time, and acceleration. The equation is:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( s \) = displacement (6.00 m)
- \( u \) = initial velocity (which we need to find)
- \( a \) = acceleration (which is \( g = 10 \, \text{m/s}^2 \))
- \( t \) = time (0.200 s)

### Step 3: Substitute Known Values
Since the ball is dropped, the initial velocity \( u \) at the start of the last 6 m is the final velocity \( v \) after falling from height \( h_1 \). We can express \( v \) in terms of \( h_1 \) using the first equation of motion:
\[ v^2 = u^2 + 2as \]
Since \( u = 0 \) when it starts falling from height \( h_1 \):
\[ v^2 = 2gh_1 \]

### Step 4: Apply the Second Equation of Motion
Now, we can apply the second equation of motion to the last 6 m:
\[ 6 = v(0.200) + \frac{1}{2}(10)(0.200^2) \]
Substituting \( v \) from the previous step:
\[ 6 = \sqrt{2gh_1}(0.200) + \frac{1}{2}(10)(0.040) \]
\[ 6 = \sqrt{20h_1}(0.200) + 1 \]

### Step 5: Rearranging the Equation
Now, we can rearrange the equation:
\[ 6 - 1 = 0.200 \sqrt{20h_1} \]
\[ 5 = 0.200 \sqrt{20h_1} \]
\[ \sqrt{20h_1} = \frac{5}{0.200} = 25 \]

### Step 6: Solve for \( h_1 \)
Now, square both sides:
\[ 20h_1 = 625 \]
\[ h_1 = \frac{625}{20} = 31.25 \, \text{m} \]

### Step 7: Find Total Height
The total height \( H \) is the sum of \( h_1 \) and the last 6 m:
\[ H = h_1 + 6 = 31.25 + 6 = 37.25 \, \text{m} \]

### Final Answer
The height from which the ball was dropped is approximately **37.25 meters**.

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