Question

If the distance travelled by a freely falling body in the last but one second of its motion is 5 m, then the last second is:

• A. \( 1^{st} \)
• B. \( 2^{nd} \)
• C. \( 3^{rd} \)
• D. \( 4^{th} \)

Hint:

For freely falling objects, use: \[ s_n = u + \frac{1}{2} g (2n - 1) \] to determine distance traveled in specific seconds.

Answer 1

The Correct Option is C

Using the equation of motion: \[ s_n = u + \frac{1}{2} g (2n - 1) \] where: - \( s_n \) = Distance traveled in \( n^{th} \) second. - \( g = 10 \) m/s\(^2\). - \( u = 0 \) (freely falling body). Step 1: Finding \( n \) \[ 5 = \frac{1}{2} \times 10 \times (2n - 1) \] \[ 5 = 5(2n - 1) \] \[ 2n - 1 = 1 \] \[ 2n = 3 \] \[ n = 3 \] Thus, the last second of motion is the third second.