At $t=0$ a $1.0\text{Kg}$ball is thrown from a tall tower with $\overrightarrow{\mathrm{v}}=\left(18\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{i}}+\left(24\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{j}}$. What is ∆U of the ball-Earth system between $t=0$and $t=6.0\text{s}$ (still free fall)?

1. The mass of the ball is$m=10\mathrm{kg}$
2. The velocity vector of the ball is$\overrightarrow{v}=\left(18m/s\right)i++\left(24\text{m/s}\right)\text{j}$
3. The initial time of the ball is$t_1=0\text{s}$
4. The final time of the ball is$t_2=6.\text{0s}$

The problem deals with the kinematic equation of motion in which the motion is described at constant acceleration. Use the second kinematic equation and the concept of gravitational potential energy to find the change in potential energy of the ball-earth system.

Formula is as follow:

$\Delta U=mg\Delta y$

Where, mis mass, gis an acceleration due to gravity, is displacement and is potential energy.

The ball is thrown from a tall tower. The vertical distance $\Delta y$covered by the ball within the time $t=6.0\text{s}$ and the vertical velocity of the ball is $v_{0y}=2\text{4m/s}$.

According to the second kinematical equation,

$$\begin{gathered} \Delta y=v_{0y}t-\frac{1}{2}g{t}^2 \\ \Delta y=\left(24\frac{\text{m}}{\text{s}}\times 6.0\text{s}\right)-\frac{1}{2}\times \left(9.8{\text{m/s}}^{\text{2}}\right)\times {\left(6.0\text{s}\right)}^2 \\ \Delta y=-32.\text{4m} \end{gathered}$$

The expression of the gravitational potential energy is,

$\Delta U=mg\Delta y$

$\Delta U=\text{1.0kg}\times \left({\text{9.8m/s}}^{\text{2}}\right)\times -\text{32.4m}$

role="math" localid="1663128704767" $\Delta U=-317\text{J}$

$\Delta U=-3.2\times {10}^2\text{J}$

Hence, the $\Delta U$of the ball-earth system is$\Delta U=-3.2\times {10}^2\text{J}$ .