Answer:
The speed of the flashlight at that point is 3.7 m/s
Explanation:
When an object of mass M is at a height H above the ground, the potential energy of the object is:
U = M*H*g
Where g is the gravitational acceleration, g = 9.8 m/s^2
And for an object with velocity v, the kinetic energy is:
K = (M/2)*v^2
We know that when the flashlight of mass 1kg is 0.7 meters above the ground, the potential energy is equal to the kinetic energy, then:
M = 1kg
H = 0.7m
g = 9.8 m/s^2
Replacing these in the equations, we get:
U = K
(1kg)*(0.7m)*(9.8 m/s^2) = ((1kg)/2)*v^2
As the mass factor appears in both sides, we can remove it:
(0.7 m)*(9.8 m/s^2) = (v^2)/2
Now we can multiply both sides by 2:
2*(0.7 m)*(9.8 m/s^2) = v^2
Now let's apply the square root to both sides:
√(2*(0.7 m)*(9.8 m/s^2)) = v = 3.7 m/s
