Question

a 25.0kg child jumps to the ground from a structure 1.00m high and comes to rest 0.500s after first contact with the ground. what average force is applied by the child in coming to rest

Answer 1

Approximately
`470\; {\rm N}`.

(Assuming that
`g = 9.81\; {\rm N \cdot kg^(-1)}` and that air resistance is negligible.)

Step-by-step explanation:

Under the assumptions, the acceleration of the child would be constantly
`a = g = 9.81\; {\rm N \cdot kg^(-1)} = 9.81\; {\rm m\cdot s^(-2)}` while the child was in the air.

Apply the SUVAT equation
`v^(2) - u^(2) = 2\, a\, x` to find the velocity
`v` of the child right before landing:

`\begin{aligned}v &= \sqrt{u^(2) + 2\, a\, x}\end{aligned}`, where:

• 
  `u = 0\; {\rm m\cdot s^(-1)}` is the initial velocity of the child,
• 
  `a = 9.81\; {\rm m\cdot s^(-2)}` is the vertical acceleration, and
• 
  `x = 1.00\; {\rm m}` is the vertical displacement (change in height.)

The child is at rest
`\Delta t = 0.500\; {\rm s}` after contact. During that
`0.500\; {\rm s}`, velocity would have changed by
`\Delta v = \sqrt{u^(2) + 2\, a\, x}`. Momentum of the child would have changed by
`\Delta p = m\, \Delta v`, where
`m = 25.0\; {\rm kg}` is the mass of the child.

Divide this change in momentum by the duration
`\Delta t` to find the average net force:

`\displaystyle F_{\text{net}} = (m\, \Delta v)/(\Delta t)`.

There are two forces on the child: upward normal force from the ground
`F_{\text{normal}}` and downward gravitational attraction
`m\, g` from the Earth. The resultant force on the child points upwards:

`F_{\text{net}} = F_{\text{normal}} - m\, g`.

Rearrange this equation to find the normal force on the child:

`\begin{aligned}F_{\text{normal}} &= F_{\text{net}} + m\, g \\ &= (m\, \Delta v)/(\Delta t) + m\, g \\ &= \frac{m\, \sqrt{2\, a\, x + u^(2)}}{\Delta t} + m\, g\\ &= \frac{(25.0)\, \sqrt{2\, (9.81)\, (1.00) + 0^(2)}}{0.500}\; {\rm N} + (25.0)\, (9.81)\; {\rm N}\\ &\approx 470\; {\rm N}\end{aligned}`.

This normal force from the ground on the child is the reaction to the force that the child exerted on the ground. The two forces will have the same magnitude: approximately
`470\; {\rm N}`. Hence, the child would have exerted an average force of approximately
`470\; {\rm N}\!` on the ground during that
`0.500\; {\rm s}`.