A person jumps from the roof of a house 2.8 m high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.70 m. If the mass of his torso (excluding legs) is 42 kg, find (a) his velocity just before his feet strike the ground, and (b) the average force exerted on his torso by his legs during deceleration.

In such a problem, use the kinematic relation to determine the velocity of the person just before his feet strike the ground. For deceleration, consider the downward direction as positive.

Given data:

The mass of the torso is $m=42\mathrm{kg}$.

The approximate distance is $d=0.7\mathrm{m}$.

The height from which the person jumps is $h=2.8\mathrm{m}$.

The relation to calculate the velocity is given by:

$v^2=u^2+2gh$

Here, g is the gravitational acceleration, v is the final velocity of the person, and u is the initial velocity whose value is zero.

On plugging the values in the above relation, you get:

$\begin{array}{c}{v}^2={\left(0\right)}^2+2\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)\left(2.8\mathrm{m}\right)\\ v=7.41\mathrm{m}/\mathrm{s}\end{array}$

Thus, $v=7.41\mathrm{m}/\mathrm{s}$is the required velocity.

The free body diagram is as follows:

The relation to calculate the average deceleration is given by:

$a=-\frac{v^2}{2d}$

On plugging the values in the above relation, you get:

$\begin{array}{c}a=-\left[\frac{{\left(7.41\mathrm{m}/\mathrm{s}\right)}^2}{2\left(0.7\mathrm{m}\right)}\right]\\ a=-39.2\mathrm{m}/{\mathrm{s}}^2\end{array}$

The relation to calculate the vertical forces is given by:

$\begin{array}{c}\Sigma F_v=ma\\ W-F=ma\\ mg-F=ma\end{array}$

Here, W is the weight, and F is the required average force.

On plugging the values in the above relation, you get:

$\begin{array}{c}\left(42\mathrm{kg}\right)\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)-F=\left(42\mathrm{kg}\right)\left(-39.2\mathrm{m}/{\mathrm{s}}^2\right)\\ F=2058.42\mathrm{N}\end{array}$

Thus, $F=2058.42\mathrm{N}$is the average force.