Question

Suppose a cannon is sitting on top of a 50.0 m high hill and a 5.00 kg cannon ball is fired with a velocity of 30.0 m/s at some unknown angle. What is the velocity of the cannon ball when it strikes the earth?

Answer 1

The final velocity of the cannon ball when it strikes the earth is 43.36 m/s.

The given parameters:

• Height of the hill, h = 50 m
• Mass of the cannon, m = 5 kg
• Velocity of the ball, v = 30 m/s

The final velocity of the cannon ball when it strikes the earth is calculated by applying the principle of conservation of energy as follows;

$P.E_i + K.E_i = P.E_f + K.E_f\\\\mgh_i + \frac{1}{2} mv_i^2 = mgh_f + \frac{1}{2} mv_f^2\\\\gh_i + \frac{1}{2} v_i^2 = g(0) + \frac{1}{2} v_f^2\\\\2gh_i + v_i^2 = v_f^2\\\\v_f = \sqrt{2gh_i + v_i^2 } \\\\v_f = \sqrt{(2 \times 9.8 \times 50) \ \ + \ \ 30^2} \\\\v_f = 43.36 \ m/s$

Thus, the final velocity of the cannon ball when it strikes the earth is 43.36 m/s.

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