Respuesta :
Answer:
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%
The player spends 78.7% of the range in the upper half of the jump.
Given that the initial velocity of the player v₀=6.5m/s at an angle θ = 36°.
Resolving the velocity components:
The horizontal component of velocity [tex]v_x=v_0cos36=5.26m/s[/tex]
The verticle component of velocity [tex]v_y=v_0sin36=3.82m/s[/tex]
The range R of the projectile is given by:
[tex]R=\frac{v_0^2sin2\theta}{g}\\ \\R=\frac{6.5^2sin72}{9.8}\\ \\R=5.4m[/tex]
And the maximum Height H of the projectile:
[tex]H=\frac{v_y^2}{2g}\\ \\ H=\frac{(3.82)^2}{2*9.8}\\\\H=0.744m[/tex]
half-maximum height h will be:
h = H/2
h = 0.372m
Now, we need to calculate the time taken to reach the half-maximum height:
[tex]h=v_yt-\frac{1}{2}gt^2\\ \\0.372=3.82t-0.5*9.8t^2\\\\4.9t^2-3.82t+0.372=0\\\\t=0.66s\\or\\ t=0.11s[/tex]
there are two values of t since the player passes height h once while jumping up and once while falling down. So we will tale t=0.11s to calculate the range r and then double it.
[tex]r=v_xt\\r=5.26*0.11m\\r=0.578m\\R'=2r=1.15m[/tex]is
R' is the range covered in the jump below the half-maximum height.
The percentage of range covered above half-maximum height:
[tex]=\frac{R-R'}{R}\times 100\\\\=\frac{5.4-1.15}{5.4}\times 100\\\\=78.7[/tex]
Hence the player spends 78.7% of the range in the upper half.
Find out more about projectile motion:
https://brainly.com/question/11261462?referrer=searchResults
