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 * 9.8 * 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. = (1)/(2) m {v}^(2)`
• Putting the values, we get
• 
  `K.E. = (1)/(2) * 12 * 98 \\ = 6 * 98 \\ = 588J`

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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