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 solve the problem of finding the height from which the ball was dropped, we can follow these steps:

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

### Step 2: Use the equations of motion
We will use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s \) is the displacement (6.00 m),
- \( u \) is the initial velocity before crossing the last 6 m,
- \( a \) is the acceleration due to gravity (10 m/s²),
- \( t \) is the time taken (0.200 s).

### Step 3: Substitute the known values
Substituting the known values into the equation:
\[ 6 = u(0.200) + \frac{1}{2} (10)(0.200)^2 \]

### Step 4: Calculate the second term
Calculate \( \frac{1}{2} (10)(0.200)^2 \):
\[ \frac{1}{2} (10)(0.200)^2 = \frac{1}{2} (10)(0.04) = 0.2 \]

### Step 5: Rewrite the equation
Now rewrite the equation:
\[ 6 = 0.200u + 0.2 \]

### Step 6: Solve for \( u \)
Rearranging gives:
\[ 6 - 0.2 = 0.200u \]
\[ 5.8 = 0.200u \]
\[ u = \frac{5.8}{0.200} = 29 \, \text{m/s} \]

### Step 7: Find the height from which the ball was dropped
Now we need to find the total height (H) from which the ball was dropped. We will use the first equation of motion again:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) is the final velocity (29 m/s),
- \( u \) is the initial velocity (0 m/s, since it was dropped),
- \( a \) is the acceleration due to gravity (10 m/s²),
- \( s \) is the distance fallen before reaching the velocity \( v \).

### Step 8: Substitute values into the equation
Substituting the known values:
\[ (29)^2 = 0 + 2(10)s \]
\[ 841 = 20s \]

### Step 9: Solve for \( s \)
Now solve for \( s \):
\[ s = \frac{841}{20} = 42.05 \, \text{m} \]

### Step 10: Calculate the total height (H)
The total height from which the ball was dropped is:
\[ H = s + 6 = 42.05 + 6 = 48.05 \, \text{m} \]

### Final Answer
The height from which the ball was dropped is approximately **48.05 meters**.

---