Question

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?

Answer 1

3.43103482932 m/s

2.25w

Step-by-step 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

`v^2-u^2=2as\\\Rightarrow v=√(2gs+u^2)\\\Rightarrow v=√(2* 9.81* 0.6+0^2)\\\Rightarrow v=3.43103482932\ m/s`

The initial speed of the person is 3.43103482932 m/s

now s = 48 cm

`v^2-u^2=2as\\\Rightarrow a=(v^2-u^2)/(2s)\\\Rightarrow a=(3.43103482932^2-0^2)/(2* 0.48)\\\Rightarrow a=12.2625\ m/s^2`

The force is given by

`F=(w)/(g)(a+g)\\\Rightarrow F=(w)/(9.81)(12.2625+9.81)\\\Rightarrow F=2.25w`

The force is 2.25w