Question

A man of mass 70 kg jumps to a height of 0.8 m from the ground, then the momentum transferred by the ground to the man is 
(g = 10 m/s\(^{-2}\)):

• A. \(280 { kg ms}^{-1}\)
• B. \(200 { kg ms}^{-1}\)
• C. \(560 { kg ms}^{-1}\)
• D. \(400 { kg ms}^{-1}\)

Hint:

To determine the momentum change for a jump, use the kinematic equations to find the take-off speed, then multiply by mass to find the change in momentum.

Answer 1

The Correct Option is A

Step 1: Calculate the initial velocity required to reach 0.8 m. 
Using the formula \( v^2 = u^2 + 2as \) and rearranging for \( u \) when \( v = 0 \), \( a = -g \), and \( s = 0.8 { m} \):
\[ 0 = u^2 - 2 \times 10 \times 0.8 \Rightarrow u^2 = 16 \Rightarrow u = 4 { m/s}. \] Step 2: Calculate the momentum transferred. 
\[ p = m \times u = 70 \times 4 = 280 { kg ms}^{-1}. \] The momentum calculation involves the mass and initial velocity, assuming no air resistance and perfect energy conversion.