A man drops a 10 kg rock from the top of a 5 m ladder, what is its speed just before it hits the ground What is it kinetic energy when it reaches the ground? ( $g = 10 m / s^{2}$ )

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Solution

Given,
mass of rock = m = 10 kg
height = 5m

using conservation of mechanical energy
loss in potential energy = gain in kinetic energy
$K . E = m g h = 10 \times 10 \times 5 = 500 J$
$1 / 2 m \times v^{2} = 500$
$v^{2} = 100$

$v = 10 m / s$

speed of rock is 10 m/s and its kinetic energy is 500 J

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A man drops a 10 kg rock from the top of a 5 m ladder, what is its speed just before it hits the ground What is it kinetic energy when it reaches the ground? (g=10m/s2)

Given, mass of rock = m = 10 kg height = 5m using conservation of mechanical energy loss in potential energy = gain in kinetic energy K.E=mgh=10×10×5=500J 1/2m×v2=500 v2=100 v=10m/sspeed of rock is 10 m/s and its kinetic energy is 500 J

A man drops a 10 kg rock from the top of a 5 m ladder, what is its speed just before it hits the ground What is it kinetic energy when it reaches the ground? ( $g = 10 m / s^{2}$ )

Given,
mass of rock = m = 10 kg
height = 5m

using conservation of mechanical energy
loss in potential energy = gain in kinetic energy
$K . E = m g h = 10 \times 10 \times 5 = 500 J$
$1 / 2 m \times v^{2} = 500$
$v^{2} = 100$

$v = 10 m / s$

speed of rock is 10 m/s and its kinetic energy is 500 J