Question

A body dropped from top of a tower fall through 40 m during the last two seconds of its fall. The height of tower is
\((g = 10\, m/s^2)\)

• A. 60 m
• B. 45 m
• C. 80 m
• D. 50 m

Answer 1

The Correct Option is B

To solve this problem, we need to determine the total height of the tower from which the body is dropped. Given that the acceleration due to gravity \( g = 10 \, \text{m/s}^2 \), we can use the equations of motion to find the solution.

We know that the body falls 40 m in the last two seconds of its fall. Let the total time taken to fall from the top of the tower be \( T \) seconds.

The distance travelled by an object in time \( t \) under constant acceleration is given by the equation:

S = ut + \frac{1}{2}at^2

Here, \( u = 0 \) m/s (initial velocity, as the body is dropped), \( a = g = 10 \, m/s^2 \), \( t = \text{time taken} \).

We need to find the height of the tower, which is the total distance travelled by the body in \( T \) seconds:

H = \frac{1}{2}gT^2

For the last 2 seconds, we consider the displacement as follows:

The displacement for the last two seconds can be calculated as:

S_{last \, 2s} = \frac{1}{2}gT^2 - \frac{1}{2}g(T - 2)^2

Given that \( S_{last \, 2s} = 40 \, \text{m} \):

\frac{1}{2} \times 10 \times T^2 - \frac{1}{2} \times 10 \times (T - 2)^2 = 40

Simplify this equation:

5T^2 - 5(T - 2)^2 = 40

Expanding the terms:

5T^2 - 5(T^2 - 4T + 4) = 40

5T^2 - 5T^2 + 20T - 20 = 40

20T - 20 = 40

20T = 60

T = 3

Now, substitute \( T = 3 \) back into the equation for total height \( H \):

H = \frac{1}{2} \times 10 \times 3^2 = 45 \, \text{m}

Therefore, the height of the tower is 45 m. Hence, the correct answer is 45 m.