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

Question

Physics

Jagger Robbins

Yesterday, 01:59

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

+1

Answers (1)

Know the Answer?

Mariana Benitez · 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