Question

A person reaches a maximum height of 62 cm 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 50 cm. To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump.

Part A: With what initial speed does the person leave the ground to reach a height of 62 cm?

Part B: In terms of this jumper's weight W , what force does the ground exert on him or her during the jump?

I figured out the first part, but I can't figure out the second part. :(

Also,

It has to be a number? This is online homework and the answer is in the format:

F=_____W

Answer 1

V= 3.485 m/s

F= 2.24 W

Step-by-step explanation:

Given:

Height 'h'= 62cm = 0.62m

distance 'x' = 50 cm= 0.5m

a)

In order to find initial speed, we use the formula

V² = 2gh

where,

g is accalaration due to gravity i.e 9.8m/s²

V² = 2 x 9.8 x 0.62

V²= 12.152

taking square root on both sides

V = 3.485 m/s

Thus, the person leave the ground to reach a height of 62 cm at a speed of 3.485 m/s

b)

F = reaction of the ground

F = mg + ma

First is to find acceleration 'a'

So, V² = 2ax

a = V²/2x = 2gh/2x

a= gh/x

F = mg + mgh/x (∵ W=mg)

F = W + Wh/x (taking 'W' common)

F = W(1+h/x)

substituting 'h' and 'x' in the above equation,

F = (1 + 0.62/0.5)W

F = 2.24 W