A film of Jesse Owens’s famous long jump (Fig. 6–47) in the 1936 Olympics shows that his center of mass rose 1.1 m from launch point to the top of the arc. What minimum speed did he need at launch if he was traveling at at the top of the arc?

The law of conservation of energy states that energy can neither be created nor destroyed but can be converted from one form to another. The total energy of a closed system remains constant.

Let the mass of Jesse be m and his initial speed be\({v_1}\).

The height of the centerof mass above the ground is\(h = 1.1\;{\rm{m}}\).

The final speed of Jesse at the top is \({v_2} = 6.5\;{\rm{m/s}}\).

The potential energy at the surface of the earth \(\left( {P{E_{\rm{i}}}} \right)\) is zero. Jessetakes off to a height hwhere the potential energy is the maximum. Applying the law of conservation of energy at the surface and top of the arc, you get:

\(\begin{aligned}K{E_{\rm{i}}} + P{E_{\rm{i}}} &= K{E_{\rm{f}}} + P{E_{\rm{f}}}\\\frac{1}{2}mv_1^2 + 0 &= \frac{1}{2}mv_2^2 + mgh\\\frac{1}{2}mv_1^2 &= \frac{1}{2}mv_2^2 + mgh\end{aligned}\) … (i)

Rearranging equation (i) gives the launch speed of Jesse before attempting the long jump.

\(\begin{aligned}\frac{1}{2}v_1^2 &= \frac{1}{2}v_2^2 + gh\\{v_1} &= \sqrt {v_2^2 + 2gh} \\ &= \sqrt {{{\left( {6.5\;{\rm{m/s}}} \right)}^2} + 2\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {1.1\;{\rm{m}}} \right)} \\ &\approx 8\;{\rm{m/s}}\end{aligned}\)

Thus, Jesse Owen has a minimum launch speed of \(8\;{\rm{m/s}}\).