Question

A 1kg cannon ball.is fired horizontally with an initial velocity of 5m/s. If the cannon was atop a wall 20m above the ground, what is the total
change in KE?

Answer 1

Answer:

Ek = 196.2 [J]

Explanation:

The question concerns the KE kinetic energy.

That is, we must find the kinetic energy at the moment the cannon is fired and the kinetic energy of when the ball hits the ground after having fallen 20 meters.

At the moment when the ball is fired it is 20 meters above ground level. If the ground level is taken as the reference level of potential energy, where it is equal to zero, in this way when the ball is at the highest (20 meters) you have the maximum potential energy.

In this way, the energy in the initial state is equal to the sum of the kinetic energy plus the potential energy. As the energy is conserved this same energy will be present when the ball hits the ground, where the potential energy is zero and will have only kinetic energy.

$E_{1}=E_{2}\\E_{k1}+E_{p1}=E_{k2}\\\frac{1}{2} *m*v^{2} +m*g*h=E_{k2}\\E_{k2}=0.5*1*(5)^{2} +1*9.81*20\\E_{k2}=208.7[J]$

The kinetic energy in the initial state can be easily calculated by means of the following equation.

$E_{k1}=\frac{1}{2} *m*v^{2}\\E_{k1}=0.5*1*(5)^{2}\\E_{k1}=12.5 [J]$

Therefore the change in KE

$E_{k} = 208.7 - 12.5\\E_{k} = 196.2 [J]$