Answer:
The cannon ball hits the ground with a speed of 76.35 m/s.
Explanation:
We shall use the conservation of energy principle to solve the problem
The initial energy of the cannon ball is the sum of the potential and the kinetic energies.
We know that potential energy = [tex]mgh[/tex]
Applying the given values we obtain the initial potential energy as
[tex]P.E=m\times g\times 18.0[/tex]
Similarly the initial kinetic energy of the ball equals
[tex]K.E_{initial}=\frac{1}{2}mv_{o}^{2}[/tex]
Applying values we get
[tex]K.E_{initial}=\frac{1}{2}m(74.0)^{2}[/tex]
Thus the initial energy is thus [tex]E_{initial}=m\times g\times 18.0+\frac{1}{2}\times m\times (74.0)^{2}[/tex]
Now upon hitting the ground the only energy that the cannon ball will posses is kinetic energy since the potential energy of any object upon touching the surface of earth equals zero
Thus we have
[tex]Energy_{final}=\frac{1}{2}mv_{f}^{2}[/tex]
Equating initial and final energies we get
[tex]m\times g\times 18+\frac{1}{2}\times m\times (74.0)^{2}=\frac{1}{2}\times m\times v_{f}^{2}\\\\v_{f}^{2}=36g+(74.0)^{2}\\\\\therefore v_f=\sqrt{36\times 9.81+(74.0)^{2}}\\\\v_f=76.35m/s[/tex]
