- Mass of the bullet (m1) = 8g = 0.008 Kg
- Velocity of the bullet (v1) = 420 m/s
- We know, kinetic energy =
![(1)/(2)m {v}^(2)](https://img.qammunity.org/2022/formulas/physics/high-school/nclfq6szm11xtcrq202tjgk91y776vchs6.png)
- Therefore, the kinetic energy of the bullet
![= (1)/(2) * 0.008 * 420J \\ = 1.68J](https://img.qammunity.org/2022/formulas/physics/high-school/7vkd2j4xbuoxwcq7f217jgvt40831ztdsc.png)
- So, the kinetic energy of the bullet is 1.68 J. It is said that the man's kinetic energy should be same as that of the bullet.
- Mass of the man (m2) = 75 Kg
- Let the velocity of the man be v2.
- Therefore,
![1.68J = (1)/(2) * 75 \: kg * v2 \\ = > v2 = (1.68 * 2)/(75) m {s}^( - 1) \\ = 0.0448 \: m \: {s}^( - 1)](https://img.qammunity.org/2022/formulas/physics/high-school/phyp5ya6rn80hoj5zy4hr5fe2joqmncgwy.png)
Answer:
0.0448 m/s
Hope you could get an idea from here.
Doubt clarification - use comment section.