Question

A cannonball is fired with initial speed v0 at an angle 30° above the horizontal from a height of 40 m above the ground. The projectile strikes the ground with a speed of 1.2v0. Find v0. (Ignore any effects due to air resistance.)

Answer 1

Answer:

$v_{0}=42.23m/s$

Explanation:

We shall use conservation of energy to solve this problem

Initial energy of the ball = Kinetic energy + potential energy

$Energy_{Initial}=\frac{1}{2}mv_{0}^{2}+mgh$

When the ball finally hit's ground it's potential energy becomes zero thus final energy becomes

$Energy_{final}=\frac{1}{2}mv_{f}^{2}+0\\\\Energy_{final}=\frac{1}{2}m(1.2v_{0})^{2}\\\\Energy_{final}=0.72mv_{0}^{2}$

Equating both the energies and applying the given values we get

$\\\\0.5mv_{0}^{2}+mgh=0.72mv_{0}^{2}\\\\0.22v_{0}^{2}=gh\\\\v_{0}=\sqrt{\frac{g\times 40}{0.22}}\\\\v_{0}=42.23m/s$