Question:(II) In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole. With what minimum speed must the athlete leave the ground in order to lift his center of mass 2.10 m and cross the bar with a speed of\({\bf{0}}.{\bf{5 m/s}}\).

In the conservation of energy, the energy at the beginning of the jump equals the total energy after the jump.

The energy at the start and after the jump is the sum of the athlete's kinetic and potential energies.

There exists some kinetic energy when the athlete jumps. Also, some potential energy exists when the athlete jumps at some height. The kinetic energy also exists while the athlete crosses the bar.

The given data can be listed below as:

• The final height of the athlete is\({h_f} = 2.10{\rm{ m}}\).
• The final speed of the athlete while crossing the bar is\({v_f} = 0.50{\rm{ m/s}}\).
• The acceleration due to gravity is \(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).

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

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

\(0 + \frac{1}{2}mv_i^2 = mg{h_f} + \frac{1}{2}mv_f^2\) … (i)

Here,\({E_i}\)is the total energy at the ground at the start of the jump, and\({E_f}\)is the total energy at the top after achieving some height.

The initial height\({h_i}\)is equal to zero,\({h_f}\)is the final height at the top,\({v_i}\)is the initial velocity at the ground, and\({v_f}\)is the final velocity at the top while crossing the bar.

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 top, and \(K.{E_f}\) is the final kinetic energy while crossing the bar.

From equation (i), the minimum velocity of the athlete can be expressed as:

\(\begin{array}{c}\frac{1}{2}mv_i^2 = mg{h_f} + \frac{1}{2}mv_f^2\\\frac{1}{2}v_i^2 = g{h_f} + \frac{1}{2}v_f^2\\v_i^2 = 2g{h_f} + v_f^2\\{v_i} = \sqrt {\left( {2g{h_f} + v_f^2} \right)} \end{array}\)

Substitute the values in the above expression.

\(\begin{array}{c}{v_i} = \sqrt {\left( {2g{h_f} + v_f^2} \right)} \\ = \sqrt {\left( {2 \times 9.81{\rm{ m/}}{{\rm{s}}^2} \times 2.10{\rm{ m}} + {{\left( {0.50\;{\rm{m/s}}} \right)}^2}} \right)} \\ = \sqrt {41.452} {\rm{ m/s}}\\ = 6.44{\rm{ m/s}}\end{array}\)

Thus, the minimum velocity of the athlete is 6.44 m/s.