An average person can reach a maximum height of about 60 cm when jumping straight up from a crouched position. During the jump itself, the person's body from the knees up typically rises a distance of around 50 cm. To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump.
I need to answer this:
a) With what initial speed does the person leave the ground to reach a height of 60 cm?
and
b) In terms of this jumper's weight W , what force does the ground exert on him or her during the jump?

Explanation:

Height attained by body = 50 cm

= .5 m

Initial velocity = u

v² = u² - 2gh

0 = u² - 2gh

u² = 2 x 9.8 x .5

u = 3.13 m /s

During the initial period , the muscle stretches by around 10 cm during which force by ground reacts on the body and gives acceleration to achieve velocity of 3.13 m/s from zero .

v² = u² + 2as

3.13² = 0 + 2 a x .10

a = 49  m/s²

reaction by ground R

Net force

R-mg = ma

R= m ( g +a )

= mg + ma

=W + (W/g) x a

W ( 1 + a / g )

= W ( 1 + 49 / 9.8 )

= 6W

the answer is b ) square root of 5

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer