Answer:
[tex]u = 7m {s}^{ - 1} [/tex]
Explanation:
We know that when we don't have air friction on a free fall the mechanical energy (I will symbololize it with ME) is equal everywhere. So we have:
[tex]me(1) = me(2)[/tex]
where me(1) is mechanical energy while on h=10m
and me(2) is mechanical energy while on the ground
Ek(1) + DynamicE(1) = Ek(2) + DynamicE(2)
Ek(1) is equal to zero since an object that has reached its max height has a speed equal to zero.
DynamicE(2) is equal to zero since it's touching the ground
Using that info we have
[tex]m \times g \times h = \frac{1}{2} \times m \times u {}^{2} \\ [/tex]
we divide both sides of the equation with mass to make the math easier.
[tex]9.8 \times 10 = \frac{1}{2} \times u {}^{2} \\ \frac{98}{2} = u {}^{2} \\ u { }^{2} = 49 \\ u = 7[/tex]
