contestada

A basketball star covers 2.70 m horizontally in a jump to dunk the ball (see figure). His motion through space can be modeled precisely as that of a particle at his center of mass. His center of mass is at elevation 1.02 m when he leaves the floor. It reaches a maximum height of 1.80 m above the floor and is at elevation 0.950 m when he touches down again.(a) Determine his time of flight (his "hang time") s(b) Determine his horizontal velocity at the instant of takeoff. m/s(c) Determine his vertical velocity at the instant of takeoff. m/s(d) Determine his takeoff angle above the horizontal(e) For comparison, determine the hang time of a whitetail deer making a jump (see figure above) with center-of-mass elevations yi = 1.20 m, ymax = 2.50 m, and yf = 0.720 m

Respuesta :

Answer:

See full explanation below for answer

Explanation:

Ok to do this, let's solve for parts and explain a little bit the procedure:

a) The time of flight would be the time which he rises and then fall, so the equation to use is:

t = √2h/g

t1 = √2 * (1.8 - 1.02) / 9.8 = 0.399 s

t2 = √2 * (1.8 - 0.95) / 9.8 = 0.416 s

Therefore, the time of flight:

0.399 + 0.416 = 0.815 s

use the formula: Vx = x/t. Solving for Vx:

Vx = 2.70 / 0.815 = 3.313 m/s  

c) using the following formula

y = yo + Vy*t - 1/2gt^2

Solving:

Vy = (y - yo + 1/2gt^2) / t

Vy = 0.950 - 1.02 + (9.8*0.815^2)/2 / 0.815

Vy = -0.07 + 3.25 / 0.815

Vy = 3.902 m/s  

d) Use the formula angle = arctg (Vy/Vx)  

angle = arctg (3.902/3.313) = 49.67  

t = √2h/g

t1 = √2 * (2.5 - 1.2) / 9.8 = 0.515 s

t2 = √2 * (2.5 - 0.72) / 9.8 = 0.603 s

Therefore, the time of hang:

0.515 + 0.603 = 1.118 s

The answers are:

(a) the time of flight is 0.815s

(b) the horizontal velocity is 3.313m/s

(c) the vertical velocity is 3.902m/s

(d) the angle of takeoff is 49.67°

Given that the center of mass is initially at 1.02 m and then reaches a maximum height of 1.8m and when the basketball star touches the ground the center of mass is at 0.95m

(a) The time of flight will be:

[tex]t = \sqrt{ 2h/g}\\\\t_1 = \sqrt{2 * (1.8 - 1.02) / 9.8 }= 0.399 s\\\\t_2 = \sqrt{ 2 * (1.8 - 0.95) / 9.8} = 0.416 s[/tex]

where [tex]t_1[/tex] is the time when he jumps up and [tex]t_2[/tex] is the time when he comes back down to the ground

The time of flight t = t₁ + t₂

t = 0.399 + 0.416

t = 0.815 s

(b) the horizontal velocity

[tex]V_x = x/t\\\\V_x = 2.70 / 0.815 \\\\ V_x=3.313 m/s[/tex]

(c)from the second equation of motion:

[tex]y = y_o + V_y*t - (1/2)gt^2\\\\V_y = (y - y_o + (1/2)gt^2) / t\\\\V_y = 0.950 - 1.02 + \frac{(9.8*0.815^2)/2} {0.815}\\\\Vy = -0.07 + 3.25 / 0.815\\\\Vy = 3.902 m/s[/tex]

(d) the angle of takeoff:

[tex]\alpha =tan^{-1}(Vy/Vx)\\\\\alpha = tan^{-1}(3.902/3.313) \\\\\alpha = 49.67[/tex]

Now, for comparison

the time of flight in the second case will be

[tex]t = \sqrt{2h/g}\\\\t_1 = \sqrt{2 * (2.5 - 1.2) / 9.8} = 0.515 s\\\\t_2 = \sqrt{2 * (2.5 - 0.72) / 9.8} = 0.603 s[/tex]

Therefore, the time of flight:

t = 0.515 + 0.603

t = 1.118 s which is larger than the original case.

Learn more:

https://brainly.com/question/11678836?referrer=searchResults

Otras preguntas