A 50 gball is thrown from a window with an initial velocity of 8.0 m/sat an angle of$\mathbf{30}\mathbf{°}$above the horizontal. Using energy methods, determine (a) the kinetic energy of the ball at the top of its flight and (b) its speed when it is 3.0 mbelow the window. (c)Does the answer to part (b) depend on either (c) the mass of the ball or (d) the initial angle?

a) Mass of the ball, m = 50 g

b) Initial velocity of the ball,$v_i=8m/s$

c) Angle of inclination,$\theta =30°$

d) Distance below the window, x = 3.0 m

We use the concept of kinetic energy. First, we resolve the components of initial velocity and then we use them in the equation. At the top, the vertical component of initial velocity is zero, so there is only a horizontal component; using that, we can find the kinetic energy at top of the flight. From the equation of conservation of energy, we can find the final velocity at the given height. We can also check whether the answers are dependent on mass or angle.

Formulae:

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

The kinetic energy of the body, $KE=\frac{1}{2}m{v}^2$ (ii)

The ball is thrown at an angle,$\theta =30°$so it has two components of velocity, x and y components.

At the top of the flight y component of initial velocity becomes zero, so there is only x component that is given by:

$$\begin{gathered} V_{ix}=V_i\cos \theta \\ =8\left(\cos 30°\right) \\ =6.9282m/s^2 \end{gathered}$$

So the kinetic energy at the top of the flight is given using equation (ii) as follows:

$$\begin{gathered} K.E=0.5\left(0.05\right){\left(6.9282\right)}^2 \\ =1.2J \end{gathered}$$

Hence, the value of the kinetic energy is 1.2 J .

Initial kinetic energy of the ball will divide into final kinetic energy and potential energy at 3.0m height below the window. Thus, the final velocity of the ball is given using equations (i) and (ii) as:

$$\begin{gathered} \frac{1}{2}m{V}_i^2=\frac{1}{2}m{V}_f^2-mgh \\ 0.5V_i^2=0.5V_f^2-gh \\ 0.5{\left(8\right)}^2=0.5\left(V_f^2\right)-\left(9.8\right)\left(3.0\right) \\ 32=0.5V_f^2-29.4 \\ {V}_f^2=\frac{61.4}{0.5} \\ =122.8 \\ {V}_f=11.08 \\ \approx 11.1m/s \end{gathered}$$

Hence, the speed of the ball below 3.0m is 11.1 m/s .

From equation (1) we can see that the velocity of theball doesn’t depend on the mass of the ball.

So the answer of b) doesn’t depend on mass.

From equation (1) we can see that velocity doesn’t depend on angle. So, the answer is that b) doesn’t depend on the angle.