Question

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?

Answer 1

The initial speed at which the person leaves the ground to reach a height of 60 cm is approximately 5.49 m/s. The force exerted by the ground on the person during the jump is approximately 784 N.

Step-by-step explanation:

To determine the initial speed at which the person leaves the ground, we can use the equations of motion. The vertical displacement of the person, from the starting position to the highest point of the jump, is equal to the maximum height of 60 cm. We can use the formula:

Δy = (v₀² - v²) / (2g)

where Δy is the displacement, v₀ is the initial speed, v is the final speed (zero at the highest point), and g is the acceleration due to gravity. Plugging in the values, we have:

60 cm = (v₀² - 0) / (2 * 9.8 m/s²)

Simplifying the equation, we can convert the displacement to meters:

0.6 m = (v₀²) / (19.6 m/s²)

Using algebra, we can find the value of v₀:

v₀ ≈ 5.49 m/s

To calculate the force the ground exerts on the person during the jump, we use Newton's second law, which states that force is equal to mass multiplied by acceleration. Since the person's weight is equivalent to the force of gravity acting on them, we can calculate the force using the formula:

F = mg

where F is the force, m is the mass, and g is the acceleration due to gravity. Plugging in the values, we have:

F = (80 kg)(9.8 m/s²)

F ≈ 784 N

Answer 2

Step-by-step 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