A body is dropped from the top of a tower and it covers a 80 m distance in last two seconds of its journey. Find out `[g=10 ms^(-2)]`
(i) height of the tower
(ii) time taken by body to reach the ground
(iii) the velocity of body when it hits the ground

Answer

To solve the problem, we will use the equations of motion under uniform acceleration due to gravity. The given data is:

- Distance covered in the last 2 seconds, \( s = 80 \, m \)
- Acceleration due to gravity, \( g = 10 \, m/s^2 \)

Let \( t \) be the total time taken to reach the ground.

### Step 1: Calculate the distance covered in the last 2 seconds

The distance covered in the last 2 seconds can be calculated using the formula:

\[ s = u t + \frac{1}{2} g t^2 \]

Where:
- \( u \) is the initial velocity (which is 0 since the body is dropped),
- \( g \) is the acceleration due to gravity,
- \( t \) is the total time of fall.

The distance covered in the last 2 seconds can also be expressed as:

\[ s = h - h_t \]

Where:
- \( h \) is the total height of the tower,
- \( h_t \) is the height covered in the first \( t-2 \) seconds.

Using the equation for the distance covered in \( t \) seconds:

\[ h = \frac{1}{2} g t^2 \]

And for the distance covered in \( t-2 \) seconds:

\[ h_t = \frac{1}{2} g (t-2)^2 \]

Thus, we have:

\[ 80 = h - h_t \]

Substituting the equations for \( h \) and \( h_t \):

\[ 80 = \frac{1}{2} g t^2 - \frac{1}{2} g (t-2)^2 \]

### Step 2: Simplify the equation

Substituting \( g = 10 \, m/s^2 \):

\[ 80 = \frac{1}{2} \cdot 10 t^2 - \frac{1}{2} \cdot 10 (t-2)^2 \]

This simplifies to:

\[ 80 = 5 t^2 - 5 (t^2 - 4t + 4) \]

Expanding the equation:

\[ 80 = 5 t^2 - 5 t^2 + 20t - 20 \]

This simplifies to:

\[ 80 = 20t - 20 \]

### Step 3: Solve for \( t \)

Adding 20 to both sides:

\[ 100 = 20t \]

Dividing by 20:

\[ t = 5 \, seconds \]

### Step 4: Calculate the height of the tower

Now substituting \( t \) back into the equation for height \( h \):

\[ h = \frac{1}{2} g t^2 = \frac{1}{2} \cdot 10 \cdot (5)^2 \]

Calculating:

\[ h = \frac{1}{2} \cdot 10 \cdot 25 = 125 \, m \]

### Step 5: Calculate the velocity of the body when it hits the ground

Using the formula for final velocity:

\[ v = u + gt \]

Substituting \( u = 0 \):

\[ v = 0 + 10 \cdot 5 = 50 \, m/s \]

### Final Answers

(i) Height of the tower: \( 125 \, m \)
(ii) Time taken by body to reach the ground: \( 5 \, seconds \)
(iii) Velocity of body when it hits the ground: \( 50 \, m/s \)

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