- Mass of the toy (m) = 12 Kg
- Acceleration due to gravity (g) = 9.8 m/s^2
- Height (h) = 5 m
- Initial velocity (u) = 0
- Let the final velocity be v.
- We know, for a freely falling body,
- [tex] {v}^{2} = {u}^{2} + 2as[/tex]
- Putting the values in the above equation, we get
- [tex] {v}^{2} = {(0)}^{2} + 2 \times 9.8 \times 5 \\ = > {v}^{2} = 98 ( {m/s)}^{2} \\ [/tex]
- Let the kinetic energy of the toy just before it hits the ground be K.E.
- We know,
- [tex]K.E. = \frac{1}{2} m {v}^{2} [/tex]
- Putting the values, we get
- [tex]K.E. = \frac{1}{2} \times 12 \times 98 \\ = 6 \times 98 \\ = 588J[/tex]
Answer:
The kinetic energy of the toy just before the toy hits the ground will be 588 J.
Hope you could get an idea from here.
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