Respuesta :

Answer:

distance fallen = 122.5 m

speed = 49 m/s

Explanation:

Δd = vi*t - 0.5at^2

Δd = 0 - 0.5(9.8)(5^2)

Δd = -122.5 (i.e. 122.5m down)

a = (vf - vi) / t

9.8 = (vf - 0) / 5

vf = 49

Lanuel

The distance and final velocity of the ball are 122.5 meters and 49 m/s respectively

Given the following data:

  • Initial velocity, U = 0 m/s (since the ball is falling from rest).
  • Time, t = 5 seconds.
  • We know that acceleration due to gravity (a) for a ball in free fall is equal to 9.8 meter per seconds square.

a. To find how far (distance) the ball fell, we would use the second equation of motion.

Mathematically, the second equation of motion is given by the formula;

[tex]S = ut + \frac{1}{2} at^2\\\\S = 0(5) + \frac{1}{2} (9.8)(5^2)\\\\S = 0 + 4.9(25)[/tex]

Distance, S = 122.5 meters.

b. To determine its final  velocity at the end of 5 seconds by using the first equation of motion:

[tex]V = U + at\\\\V = 0 + 9.8(5)[/tex]

Final velocity, V = 49 m/s.

Therefore, the distance and final velocity of the ball are 122.5 meters and 49 m/s respectively.

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