Answer:
22.1 m/s
Explanation:
We can solve the problem by using the conservation of energy. The mechanical energy is equal to the sum of kinetic energy and potential energy, and it is constant:
[tex]E=K+U=\frac{1}{2}mv^2+mgh[/tex]
where
m is the mass
v is the velocity
g is the gravitational acceleration
h is the heigth of the object
At the moment when the object is dropped, the velocity is still zero, so K=0 and all the energy is gravitational potential energy:
[tex]E=U=mgh=(50 kg)(9.8 m/s^2)(25 m)=12250 J[/tex]
When the object reaches the ground, the height becomes zero: h=0, so the gravitational potential energy is zero and all the energy is now kinetic energy:
[tex]E=K=\frac{1}{2}mv^2[/tex]
So, we can find the velocity:
[tex]v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(12250 J)}{50 kg}}=22.1 m/s[/tex]
