Question

Force During a Jump.

An average person can reach a maximum height of about 60 cm when jumping straight up. During the jump itself, the person’s body from the knees up typically rises a distance of around 50 cm.

(a) With what initial speed does the person leave the ground to reach a height of 60 cm?

V(final) = V(initial) +2a(delta{y})
0 =V(initial) + 2(-9.8m/s^2)(.60m)

V(initial) = 3.4 m/s

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

V(final) = V(initial) +2a(delta{y})
(3.4m/s)^ = 0 + 2a(.05m)

a = 11.6 m/s^2

n-w = ma

n = w( 1 + a/g)

n = w( 1 + 11.6m/s^2/9.8m/s^2)

n=2.2w

Answer 1

So in my calculation the following are the answers or the result to your calculation:
a.initial Velocity is 11.76m/s
b. a = 11.sm/s^2

I hope you are satisfied with my answer and feel free to ask for more clarifications and questions. Have a nice day