A person reaches a maximum height of 65 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 41 cm. 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) In terms of this jumper's weight W , what force does the ground exert on him or her during the jump?

a) v = 3.57 m/s

b) N = 2.59 W

Explanation:

a) Maximum height, h = 65 cm = 0.65 m

The speed of the person at which he leaves the ground can be calculated by the equation of motion:

u = 0 m/s

b)

The displacement of the person when he jumps, S = 41 cm = 0.41 m

To calculate the acceleration of the person:

The force exerted by the person can be given as:

F = N - mg

N = Normal reaction ( the ground's force on him)

mg = W ( Weight of the person)

F = ma

ma = N - mg

Divide through by mg

muscle endurance

explanation:

i can do 50 in a row ( but i get so tired after that)

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.