A1.50 kgsnowball is shot upward at an angle of$\mathbf{34}\mathbf{.}\mathbf{0}\mathbf{°}$to the horizontal with an initial speed of20.0 m/s.

1. What is its initial kinetic energy?
2. By how much does the gravitational potential energy of the snowball–Earth system change as the snowball moves from the launch point to the point of maximum height?
3. What is that maximum height?

The mass of the snowball, m=1.50 kg

The angle of inclination,$\theta =34°$

The horizontal speed, v=20 m/s

Here we have to use the basic formula for kinetic energy, which is half of product of mass and square of velocity. Then use conservation of energy to find the gravitational energy.

Formula:

$KE=\frac{1}{2}m{v}^2$

Initial kinetic energy is calculated as follows:

$$\begin{gathered} KE=\frac{1}{2}m{v}^2 \\ =\frac{1}{2}\times 1.50kg\times {\left(20m/s\right)}^2 \\ =300J \end{gathered}$$

Hence, the initial kinetic energy is 300 J.

Calculateby how much does the gravitational potential energy of the snowball–Earth system change.

The ball is going up, which works against gravity.

So, the change in gravitational potential energyis a changein kinetic energy.

$$\begin{gathered} ∆U_g=K_i-K_f \\ =300J-\frac{1}{2}\times 1.5\times {\left[\left(20\cos 34°\right)m/s\right]}^2 \\ =300J-0.75\times \left[{\left(20\times 0.829\right)}^2\right]J \\ =300J-206.2J \\ ∆U_g=93.8J \end{gathered}$$

Hence, the gravitational potential energy at maximum height is 93.8 J.

The change in potential energy is define by usg following formula.

$∆U_g=mgh$

Here, g is the gravity having a value $9.8m/s^2$, m is the mass, and h is the maximum height.

Rearrange the above equation for height as below.

$h=\frac{∆U_g}{gm}$

Substitute known values in the above equation.

$$\begin{gathered} h=\frac{93.8J}{9.8m/s^2\times 1.5kg} \\ =6.38m \end{gathered}$$

Hence, the maximum height is 6.38 m.