A body is released from the top of a tower. In the last second of its free fall from rest, it covers the lower half of the height of the tower. Then the time of flight of the body is

Solution:

To solve this problem, we can apply the equations of motion for free fall.

Let the height of the tower be h, and let the time of flight be t. The body is released from rest, so its initial velocity is u = 0.

The distance covered by a freely falling object from rest is given by the equation:

h = (1/2) g t²

where:

• h is the height,
• g is the acceleration due to gravity (≈ 9.8 m/s²),
• t is the time of flight.

In the last second of its fall, the body covers half of the total distance. The distance fallen in the last second is h/2.

The distance fallen in the last second is the difference between the distance fallen at time t and at time t-1:

Distance in last second = (1/2) g t² - (1/2) g (t-1)²

This distance should equal h/2, so:

(1/2) g (t² - (t-1)²) = h/2

Expanding and simplifying:

(1/2) g (t² - (t² - 2t + 1)) = h

After simplifying:

2t - 1 = t²

Solving this quadratic equation will give us the time t.

Answer: b) 3.414 sec