91Ó°ÊÓ

The mass of the water balloon is, m = 1.50 kg

The initial speed of the water balloon is,$v_i=3m/s$

As the ball is launched from its initial position, using the formula for kinetic energy, we can find its kinetic energy when it is launched. As it reaches its full ascent, its kinetic energy will be converted to potential energy which will be called the work done by the gravitational force on the balloon. As we know the gravitational potential at the initial and final position, we can find the change in the gravitational potential. Using the equation for kinematics we can find the maximum height reached by the balloon.

Formulae:

The potential energy at a height, PE = mgh (1)

The kinetic energy of the body, $\mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2$ (2)

The change in the potential energy, $∆PE=P{E}_{final}-P{E}_{initial}$ (3)

The work done by an applied force, $W=Force×displacement$ (4)

The third equation of the kinematic equation,

$v_f^2=v_i^2+2ax$ (5)

Kinetic energy at the launch point can be given using equation (2) as follows:

$$\begin{gathered} K{E}_{initial}=\frac{1}{2}×1.50kg×{\left(3.00m/s\right)}^2 \\ =6.75kg·m^2/s^2\left(\frac{1J}{1kg·m^2/s^2}\right) \\ =6.75J \end{gathered}$$

Hence, the value of the kinetic energy is 6.75 J.

When it will reach the maximum height, its final velocity ($v_f=0$) will be zero, thus the distance traveled by the balloon can be given using equation (5):

$$\begin{gathered} 0m/s={\left(3.0m/s\right)}^2-2×9.8m/s^2×\left(-x\right) \\ 9m^2/s^2=-19.6m/s^2×x \\ x=-\frac{9m^2/s^2}{19.6m/s^2} \\ x=-0.46m \end{gathered}$$

The force of gravitation on the balloon can be given as:

$$\begin{gathered} F_{Gravitation}=mg \\ =1.50kg×9.8m/s^2 \\ =14.7kg·m/s^2\left(\frac{1N}{1kg·m/s^2}\right) \\ =14.7N \end{gathered}$$

Now, the work done by the force can be given using equation (4) as:

$$\begin{gathered} W=14.7N×\left(-0.46m\right) \\ =-6.75N·m\left(\frac{1J}{1N·m}\right) \\ =-6.75J \end{gathered}$$

Hence, the value of the work done is -6.75 J.

When the ball is at the launch position, h = 0 , thus the initial potential energy using equation (1) is given as:

${\mathrm{PE}}_{\mathrm{initial}}=0J$

When the ball is at maximum height, h = 0.46 m, thus the final potential energy using equation (1) is given as:

$$\begin{gathered} P{E}_{final}=1.5kg×9.8m/s^2×0.46m \\ =6.75kg·m^2/s^2\left(\frac{1J}{1kg·m^2/s^2}\right) \\ =6.75J \end{gathered}$$

Now, the change in potential energy is given using equation (3) as follows:

$$\begin{gathered} ∆PE=P{E}_{final}-P{E}_{initial} \\ =6.75J-0J \\ =6.75J \end{gathered}$$

Hence, the value of the change in the energy is 6.75 J.

From the above calculation, it can be seen that the potential energy at maximum height, h = 0.46 m is 6.75 J.

Now at the maximum height, the gravitational potential is$P{E}_{final}=0J$

When the ball is at its initial position, its distance from the final position will be h = 0.46 m

From part (d), the change in the potential energy can be given as:

$∆PE=6.75J$

So, the initial potential energy is given using equation (3):

$$\begin{gathered} P{E}_{initial}=∆P{E}_{final}-∆PE \\ =0J-6.75J \\ =-6.75J \end{gathered}$$

Hence, the value of the potential at the launch point is -6.75 J

When it will reach the maximum height, its velocity will be zero, hence, from the part (b) calculations, the maximum height of the balloon is 0.46 m.