A person reaches a maximum height of 57 cmcm 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 48 cmcm . 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) Draw a free-body diagram of the person during the jump.

(c) In terms of this jumper’s weight w, what force does the ground exert on him or her during the jump?

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boffeemadrid boffeemadrid · Answer 1 · 2020-03-28T08:51:12+01:00

Answer:

3.43103482932 m/s

2.25w

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

w = Weight of the person

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2gs+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.6+0^2}\\\Rightarrow v=3.43103482932\ m/s[/tex]

The initial speed of the person is 3.43103482932 m/s

now s = 48 cm

[tex]v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{3.43103482932^2-0^2}{2\times 0.48}\\\Rightarrow a=12.2625\ m/s^2[/tex]

The force is given by

[tex]F=\dfrac{w}{g}(a+g)\\\Rightarrow F=\dfrac{w}{9.81}(12.2625+9.81)\\\Rightarrow F=2.25w[/tex]

The force is 2.25w