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Scorpion4ik [409]

3 years ago

6

In the vertical jump, an athlete starts from a crouch and jumps upward as high as possible. Even the best athletes spend little

more than 1.00 s in the air (their "hang time"). Treat the athlete as a particle and let ymax be his maximum height above the floor. To explain why he seems to hang in the air, calculate the ratio of the time he is above ymax/2 to the time it takes him to go from the floor to that height. Ignore air resistance.

1 answer:

horrorfan [7]3 years ago

8 0

Answer:

t_up / t_down = 6.83

Explanation:

Find:

Calculate the ratio of the time he is above y_max/2 to the time it takes him to go from the floor to that height.

Solution:

- Compute the velocity v_o at y_max:

v_i^2 = v_f^2 - 2*g*y_max

0 = v_o^2 - 2*g*y_max

v_o = sqrt (2*g*y_max)

- The total time spend by athlete above height y_max / 2 is:

y - y_o = v_o*t_up - 0.5*g*t^2_up

v_o = 0.5*g*t_up

- Equate two equations:

sqrt (2*g*y_max) = 0.5*g*t_up

t_up = 2*sqrt(2*g*y_max) / g

- The total time taken by athlete to reach height y_max / 2 from ground is:

y - y_o = v_o*t_down - 0.5*g*t^2_down

g*t_^2down - 2*v_o*t_down + y_max = 0

- Solve the quadratic and evaluate t_down:

t_down = (v_o +/- sqrt (v^2_o - g*y_max)) / g

Substitute for v_o = sqrt (2*g*y_max)

t_down = (sqrt(2g*y_max) +/- sqrt(g*y_max)) / g

- We will use the minus quantity, because we need the first part of the journey from ground from the two times he passes the height of y_max/2.

Hence,

t_down = (sqrt(2g*y_max) - sqrt(g*y_max)) / g

t_down = (sqrt(g*y_max) / g) * (sqrt(2) - 1)

- Compute the ratio t_up to t_down:

t_up / t_down = 2*sqrt(2*g*y_max) / g * g / (sqrt(g*y_max)*(sqrt(2) - 1)

= 2*sqrt(2) / (sqrt(2) - 1)

= 6.83

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1) 5.5 N

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$T-mg = m\frac{v^2}{R}$

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T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

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and re-arranging the equation, we find

$v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s$

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