Question:(II) A 0.48-kg ball is thrown with a speed of at an upward angle of 36°. (a) What is its speed at its highest point, and (b) how high does it go? (Use conservation of energy.)

A ball is thrown upward at an angle of 36⁰. The ball's velocity has two components in the x and y directions. The ball is moving with the horizontal component velocity and reaches the maximum height.

After reaching the maximum height, the ball has kinetic energy as well as potential energy.The conservation of energy can be applied to calculate the maximum height that the ball can reach.

The given data can be listed below as:

• The initial velocity of the ball is\({v_i} = 8.8{\rm{ m/s}}\).
• The angle of inclination of the ball is\(\theta = 36^\circ \).
• The acceleration due to gravity is \(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).

The diagram can be shown as:

Here, A is considered as the initial point and B is considered as the final or highest point.

While resolving the initial velocity, it can be noted that the axis with which the velocity is inclined is considered the (cosine) component of the same initial velocity. The other axis could be considered as the (sine) component of the same initial velocity.

The y-component of the initial velocity can be expressed as:

\({v_y} = {v_i}\sin \theta \)

Here,\({v_y}\)is the y-component of the initial velocity.

The x-component of the initial velocity can be expressed as:

\({v_x} = {v_i}\cos \theta \)

Here,\({v_i}\)is the initial velocity, and\({v_x}\)is the x-component of the initial velocity, which is the speed of the ball at the highest point.

Substitute the values in the above expression.

\(\begin{array}{c}{v_x} = {v_i}\cos \theta \\ = 8.8{\rm{ m/s}} \times \cos 36^\circ \\ = 7.12{\rm{ m/s}}\end{array}\)

Thus, the speed at the highest point is 7.12 m/s.

According to the conservation of energy, the energies can be written as:

\(\begin{array}{c}{E_i} = {E_f}\\P.{E_i} + K.{E_i} = P.{E_f} + K.{E_f}\\mg{h_i} + \frac{1}{2}mv_i^2 = mgh + \frac{1}{2}mv_x^2\end{array}\)

\(0 + \frac{1}{2}mv_i^2 = mgh + \frac{1}{2}mv_x^2\) … (i)

Here,\({E_i}\)is the total energy at the initial point, and\({E_f}\)is the total energy at the top after achieving a certain height.

The initial height\({h_i}\)is equal to zero,\(h\)is the final height at the top,\({v_i}\)is the initial velocity at the ground, and\({v_x}\)is the final velocity at the highest point.

The initial potential energy\(P.{E_i}\)is equal to zero,\(K.{E_i}\)is the initial kinetic energy at the ground,\(P.{E_f}\)is the final potential energy at the highest point, and\(K.{E_f}\)is the final kinetic energy at the highest point.

From equation (i), the maximum height to which the ball can travel can be expressed as:

\(\begin{array}{c}\frac{1}{2}mv_i^2 = mgh + \frac{1}{2}mv_f^2\\\frac{1}{2}v_i^2 = gh + \frac{1}{2}v_x^2\\v_i^2 = 2gh + v_x^2\\h = \frac{{v_i^2 - v_x^2}}{{2g}}\end{array}\)

Substitute the values in the above expression.

\(\begin{array}{c}h = \frac{{{{\left( {{\rm{8}}{\rm{.8 m/s}}} \right)}^2} - {{\left( {7.12{\rm{ m/s}}} \right)}^2}}}{{2 \times 9.81{\rm{ m/}}{{\rm{s}}^2}}}\\ = 1.36{\rm{ m}}\end{array}\)

Thus, the highest point to which the ball can travel is 1.36 m.