91影视

The given data can be listed below.

The maximum height of an average person is,

$\begin{array}{rcl}{\mathrm{y}}_{\mathrm{avg}}& =& 60\mathrm{cm}\\ & =& \frac{60}{100}\mathrm{m}.\\ & =& 0.60\mathrm{m}\end{array}$

The distance of rising a person from the knees is,

$\begin{array}{rcl}\mathrm{d}& =& 50\mathrm{cm}\\ & =& \frac{50}{100}\mathrm{m}.\\ & =& 0.50\mathrm{m}\end{array}$

The equation of motion is used to calculate the motion of an object in 2-dimensional, and 3-dimensional. The equation of motion can easily calculate the position, velocity, acceleration, etc.

The expression for the initial speed can be expressed as,

${\mathrm{v}}^2={\mathrm{u}}^2+2{\mathrm{gy}}_{\mathrm{avg}}$

Here, u is the initial velocity at rest, g is the acceleration due to gravity, ${\mathrm{y}}_{\mathrm{avg}}$is the maximum height of a person.

Substitute 0 for u, $9.8\mathrm{m}/{\mathrm{s}}^2$for g, and 0.6 m for ${\mathrm{y}}_{\mathrm{avg}}$in the above equation.

$\begin{array}{rcl}{\mathrm{v}}^2& =& 0+2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(0.6\mathrm{m}\right)\\ \mathrm{v}& =& \sqrt{2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(0.6\mathrm{m}\right)}\\ & =& 3.429\mathrm{m}/\mathrm{s}\end{array}$

Hence, required initial speed is $3.429\mathrm{m}/\mathrm{s}$.

The free body diagram can be expressed as,

Here, F is the upward force and W is the weight acting downward when jumping.

The expression for the acceleration using equation of motion can be expressed as,

${\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{ad}$

Here, a is the acceleration and d is the distance.

Substitute 0 for u, 0.5 m for d, $3.429\mathrm{m}/\mathrm{s}$for v in the above equation.

$$\begin{gathered} {\left(3.429\mathrm{m}/\mathrm{s}\right)}^2=0+2\mathrm{a}\left(0.5\mathrm{m}\right) \\ \mathrm{a}=\frac{{\left(3.429\mathrm{m}/\mathrm{s}\right)}^2}{2\left(0.5\mathrm{m}\right)} \\ =11.758\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the acceleration is $11.758\mathrm{m}/{\mathrm{s}}^2$.

The expression for the force according to Newton鈥檚 second law in term of weight can be expressed as,

$\begin{array}{rcl}\mathrm{F}& =& \mathrm{ma}+\mathrm{w}\\ & =& \mathrm{ma}+\mathrm{mg}\\ & =& \mathrm{m}\left(\mathrm{a}+\mathrm{g}\right)\\ & =& \frac{\mathrm{w}}{\mathrm{g}}\left(\mathrm{a}+\mathrm{g}\right)\end{array}$

Here w is the weight.

Substitute $9.8\mathrm{m}/{\mathrm{s}}^2$for g, $11.758\mathrm{m}/{\mathrm{s}}^2$for a in the above equation.

$\begin{array}{rcl}\mathrm{F}& =& \frac{\mathrm{w}}{9.8\mathrm{m}/{\mathrm{s}}^2}\left(9.8\mathrm{m}/{\mathrm{s}}^2+11.758\mathrm{m}/{\mathrm{s}}^2\right)\\ & =& 2.2\mathrm{w}\end{array}$

Hence, required force is 2.2w.