Question

A person reaches a maximum height of 65 cm when jumping straight up from a crouched position. During the jump itself, the person's body from the knees up rises a distance of around 41 cm. To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump.
A) With what initial speed does the person leave the ground to reach a height of 60 cm?B) In terms of this jumper's weight W , what force does the ground exert on him or her during the jump?

Answer 1

Answer:

a) v = 3.57 m/s

b) N = 2.59 W

Explanation:

a) Maximum height, h = 65 cm = 0.65 m

The speed of the person at which he leaves the ground can be calculated by the equation of motion:

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

u = 0 m/s

$v^{2} = 0 + 2gh\\v^{2} = 2gh\\v^{2} = 2 * 9.8 * 0.65\\v^{2} = 12.74\\v = 3.57 m/s$

b)

The displacement of the person when he jumps, S = 41 cm = 0.41 m

To calculate the acceleration of the person:

$v^{2} = u^{2} + 2aS\\u = 0 m/s\\v^{2} = 2aS\\3.57^{2} = 2 * a * 0.41\\12.74 = 0.82 a\\a = 12.74/0.82\\a = 15.54 m / s^2$

The force exerted by the person can be given as:

F = N - mg

N = Normal reaction ( the ground's force on him)

mg = W ( Weight of the person)

F = ma

ma = N - mg

Divide through by mg

$\frac{ma}{mg} = \frac{N}{mg} - \frac{mg}{mg} \\\frac{a}{g} = \frac{N}{mg} - 1\\ W = mg\\\frac{a}{g} + 1= \frac{N}{W}\\\frac{15.54}{9.8} + 1 = \frac{N}{W}\\2.59 = \frac{N}{W}\\N = 2.59W$

Answer 2

Answer:

Explanation:

We can approach this question from Newton's Law of Motion.

The first law states that an object will continue moving on a straight line until an external force is acted upon it . Now from the classical mechanics perspective ; if the net force on an object happens  to be zero, the body of such object typically implies to be at rest.

The second law of motion states that , the acceleration of an object is directly proportional to the force acting upon it but inversely proportional to the mass.

So; mathematically; we can  have:

$\sum F = ma$

Newton third law of motion states that for every action , there is equal  but an opposite reaction.

A

From the question;

The initial speed $v_i =$0

But the final speed can be determined by the using he following relation:

$v_f^2 = v_i^2 +2a \Delta x \\ \\ v_f^2 = 0+2g \Delta x \\ \\ v_f^2 =2g \Delta x \\ \\ v_f^2 = 2*9.8*0.65 \\ \\ v_f^2 = 12.74m^2/s^2 \\ \\ v_f= \sqrt{12.74m^2/s^2} \\ \\ v_f= 3.569 \ m/s$

B

To calculate the acceleration of the of the jumper ; we use the relation;

$v_f^2 = v_i^2 +2a \Delta x \\ \\ v_f^2 = 0+2a \Delta x \\ \\ v_f^2 =2a \Delta x \\ \\ a= \dfrac{v_f^2}{2 \Delta x}$

$a= \dfrac{12.74}{2*0.41}$

a = 15.533 m/s²

$\sum F = ma\\ \\ F- W$

By rearrangement and solving for F; we have:

F = ma + W

F = ma + mg

F = m(a+ g)

F = w/g (a+g)

F = w/9.8 m/s² ( 15.533 + 9.8)m/s²

F = w/9.8 m/s² (25.333)m/s²

F = 2.585 w