Exams › NEET › Physics

What will be the ratio of the distances moved by a freely falling body from rest in 4th and 5th seconds of journey?

1. (a) 4 : 5
2. (b) 7 : 9
3. (c) 16 : 25
4. (d) 1 : 1

Correct answer: (b) 7 : 9

Solution

The distance covered in the nth second of free fall is given by the formula: Sn = u + (1/2)g(2n - 1). For the 4th second, S4 = (1/2)g(2×4 - 1) = (1/2)g(7). For the 5th second, S5 = (1/2)g(2×5 - 1) = (1/2)g(9). The ratio is S4 : S5 = 7 : 9.

Related NEET Physics questions

• In the equation of motion, \( \boldsymbol{S}=\boldsymbol{u} \boldsymbol{t}+ \) \( 1 / 2 a t^{2}, \) S stands for
• A body is accelerated by applying a force of 30 N. The change in the momentum of the body after 2 sec is?
• A passenger in moving train tosses a coin which falls behind him. It means that the motion of the train is
• From the displacement-time graph shown here, find the velocity of the body as it moves from \( A \) to \( B \)
• The distance of a particle as a function of time is shown below. The graph indicates that
• Whenever an object moves with a constant speed, its distance - time graph is a

⚔️ Practice NEET Physics free + battle 1v1 →