You throw a ball of mass 160gupward (Figure 6.79). When the ball is 2mabove the ground, headed upward (the initial state), its speed is 19m/s. Later, when the ball is again 2mabove the ground, this time headed downward (the final state), its speed is 19m/s. What is the change in the kinetic energy of the ball from initial to final state?

The given data is listed as follows,

• The mass of the ball is,$m=160\mathrm{g}$
• The initial distance of the ball above the ground is,2m
• The speed of the ball initially is,$v_1=19\mathrm{m}/\mathrm{s}$
• The final distance of the ball above the ground is,2m
• The speed of the ball finally is,$v_f=19\mathrm{m}/\mathrm{s}$

The kinetic energy is described as half of the product of the mass and the square of the velocity. It can be expressed as follows,

$\mathbf{KE}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}^{\mathbf{2}}$

Here,$m$ is the mass of the object and$v$ is the velocity of the object.

The equation of the change in the kinetic energy is expressed as:

$$\begin{gathered} ∆KE=K{E}_1+K{E}_i \\ =\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2 \end{gathered}$$

Here,$m$is the mass of the ball,$v_f$is the final velocity of the ball and$v_i$is the initial velocity of the ball.

Substitute all the values in the above equation.

$$\begin{gathered} ∆KE=\frac{1}{2}\left(160g\right){\left(19\mathrm{m}/\mathrm{s}\right)}^2-\frac{1}{2}\left(160g\right){\left(19\mathrm{m}/\mathrm{s}\right)}^2 \\ =0 \end{gathered}$$

Thus, the change in the kinetic energy of the ball from the initial to the final state is 0.