jongun6k
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A football player kicks a football, from ground level, with an initial velocity of 27.0 m/s at an angle 30° above the horizontal.
a. What is the maximum height the ball attained?
b. How long did it take the ball to return to the launching height?
C. How far away did it land?​

Respuesta :

OverpoweredNoob

Answer:

See answers below

Explanation:

a.

(vf)² = (vi)² + 2gy

0 = (27sin30°)² + 2(-9.8)y

y = 36.3 m

b.

vf = vi + gt

0 = (27sin30°) + (-9.8)t

t = 2.72 s

2.72 x 2 = 5.44 s

c.

x = (vi)t + (1/2)at²

x = (27cos30°)(5.44) + 0

x = 22.66 m

Cricetus

(a) The max height is "9.3 m".

(b) Total time is "2.75 sec".

(c) Total covered distance is "63 m".

According to the question,

  • Horizontal angle = 30°
  • Initial velocity, V = 27.0 m/s

then,

  • [tex]V_x = 23 \ m/s[/tex]
  • [tex]V_{iy} = 13.5 \ m/s[/tex]

(a)

The maximum height will be:

→ [tex]V_f^2 = V_i^2+2ad[/tex]

    [tex]0 = (13.5)^2+2(-9.8).d[/tex]

    [tex]0 = 182+(-19.6)d[/tex]

    [tex]d = 9.3 \ m[/tex]

(b)

The total time taken will be:

→ [tex]V_f = -V_i[/tex]

or,

→     [tex]a = \frac{V_f-V_i}{t}[/tex]

 [tex]-9.8 = \frac{-13.5-13.5}{t}[/tex]

[tex]-9.8 \ t = - 27[/tex]

       [tex]t = \frac{-27}{-9.8}[/tex]

          [tex]= 2.75 \ sec[/tex]

(c)

The total distance covered will be:

→ [tex]V_x = \frac{d_x}{t}[/tex]

  [tex]23=\frac{d_x}{2.75}[/tex]

  [tex]d_x = 2.75\times 23[/tex]

       [tex]= 63 \ m[/tex]

Thus the approach above is appropriate.

Learn more about time taken here:

https://brainly.com/question/24618424

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