In basketball, hang is an illusion in which a player seems to weaken the gravitational acceleration while in midair. the illusion depends much on a skilled player's ability to rapidly shift the ball between hands during the flight, but it might also be supported by the longer horizontal distance the player travels in the upper part of the jump than in the lower part. if a player jumps with an initial speed of v0 = 6.50 m/s at an angle of θ0 = 36.0˚, what percent of the jump's range does the player spend in the upper half of the jump (between maximum height and half-maximum height)?

 85%

Explanation:

Let's analyze the problem and see that it is an exercise in kinematics, where we must find the horizontal scopes for the top of the jump, the total range and compare them

Let's start by calculating the total scope

      R = Vo² sin 2θ / g

      R = 6.5² sin 2 36 /9.8

      R = 4.10 m

Now we must calculate the average jump height, let's look for the maximum height and divide by two

     Vfy2 = Voy2 - 2 g Y

Let's calculate the initial velocity component, by trigonometry

     Vox = Vo cos T

     Voy = Vo sin T

     Vox = 6.5 cos 36

     Voy = 6.5 sin 36

     Vox = 5.26 m / s

     Voy = 3.82 m / s

For the maximum height Vfy = 0

     0 = Voy² - 2 g Ymax

     Ymax = Voy² / 2g

    Ymax = 3.82²/2 9.8

    Ymax = 0.745 m

Half of this height is tm = 0.745 / 2 = 0.372 m

We need to know the horizontal distance when it is at this height, for this we calculate the time it takes to reach the average height

     Y = Voy t - ½ g t²

     0.372 = 3.82 t2 - ½ 9.8 tm²

     0 = 3.82 t - 4.9 t2 - 0.372         multiply by (-1 / 4.9)

     t² - 0.78 t + 0.076 = 0

We solve the second degree equation

   tm = [0.78 + - R (0.78² - 4 1 0.076)] / 2 = [0.78 + -R (0.6084 - 0.304)] / 2

    tm = [.78 + - 0.552] / 2

   tm1 = 0.666 s

   tm2 = 0.057 s

One time is for when it is going up and the other for when it is going down, let's use the time when it is good tm2 = 0.057 s to calculate the distance traveled

    X = vox t

    Xm = Vox tm

    Xm = 5.26 0.057

    Xm = 0.2998 m

Since the vertical acceleration is constant, the distance it travels at the end of the path is the same as it travels at the beginning, so the total half-length length of twice that calculated

     Xm all = Xm + Xm

     Xm all = 0.5996 m

This is the distance from the ground to the mid-height point, the distance from this point to the maximum height is the difference with the total distance

      X upper = R - Xm all

      X upper = 4.1 -0.5996

      X upper = 3.5 m

This is the distance traveled in the super superior of the jump

We already have the total distance traveled and halfway up, let's calculate the percentage in the high jump we make a rule of proportion

     4.1 m 100%

     3.5 m --- a

      a = 3.5 / 4.1 100

  a = 85%

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