A particle is dropped from a tower 180 m high. How long does it take to reach the ground ? What is the velocity when it touches the ground ? Take `g=10m//s^(2)`.

Explanation

To solve the problem step by step, we will use the equations of motion under uniform acceleration due to gravity.

### Step 1: Identify the known values
- Height of the tower (h) = 180 m
- Initial velocity (u) = 0 m/s (since the particle is dropped)
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Calculate the final velocity (v) when the particle touches the ground
We can use the equation of motion:
\[ v^2 = u^2 + 2gh \]

Substituting the known values:
\[ v^2 = 0 + 2 \times 10 \times 180 \]
\[ v^2 = 3600 \]

Now, take the square root to find v:
\[ v = \sqrt{3600} \]
\[ v = 60 \text{ m/s} \]

### Step 3: Calculate the time (t) taken to reach the ground
We can use another equation of motion:
\[ v = u + gt \]

Substituting the known values:
\[ 60 = 0 + 10t \]
\[ 60 = 10t \]

Now, solve for t:
\[ t = \frac{60}{10} \]
\[ t = 6 \text{ seconds} \]

### Final Answers:
- Time taken to reach the ground = 6 seconds
- Final velocity when it touches the ground = 60 m/s

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