High jump and knee bending

Problem

A high jumper bends their legs before leaving the ground. Explain in five lines why bending allows them to reach greater heights, in terms of forces and impulses.

Solution

This is a qualitative problem explained by the impulse–momentum theorem. Recall that the change in momentum of a body equals the impulse of the resultant force, i.e. the product of the force and the time over which it is applied: $\vv{J} = \vv{F}\,\Delta t = \Delta \vv{p} = m\,\Delta v$

The role of bending. By bending the legs, the athlete lengthens the stretch over which the muscles push the body upward and, above all, increases the time $\Delta t$ during which the push force acts before take-off. For the same muscular force exerted, a longer contact time means a greater impulse: $\Delta t \uparrow \quad\Rightarrow\quad J = F\,\Delta t \uparrow$

Higher take-off speed. A greater impulse produces a greater change in momentum and hence a higher take-off speed $v$: $v = \frac{J}{m} = \frac{F\,\Delta t}{m}$

Height scales as $v^2$. Once airborne, the body is in free fall and rises until the vertical velocity vanishes. From the kinematics of uniformly accelerated motion, the height reached is: $h = \frac{v^2}{2g}$

Since $h$ depends on the square of the take-off speed, even a modest increase in $v$ gained from bending translates into a significant gain in height.

$\ev{\text{Bending} \Rightarrow \Delta t \uparrow \Rightarrow J \uparrow \Rightarrow v \uparrow \Rightarrow h \propto v^2 \uparrow}$

Links

Topics: Dynamics Concepts: Impulse · Newton’s second law