Question

A 12.00 kg toy falls from a height of 5.00 m. What will the kinetic energy of the toy be just before the toy hits the ground?

Answer 1

• 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,
• ${v}^{2} = {u}^{2} + 2as$
• Putting the values in the above equation, we get
• ${v}^{2} = {(0)}^{2} + 2 \times 9.8 \times 5 \\ = > {v}^{2} = 98 ( {m/s)}^{2}$
• Let the kinetic energy of the toy just before it hits the ground be K.E.
• We know,
• $K.E. = \frac{1}{2} m {v}^{2}$
• Putting the values, we get
• $K.E. = \frac{1}{2} \times 12 \times 98 \\ = 6 \times 98 \\ = 588J$

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.

Doubt clarification - use comment section.