Question

a coin is dropped from a 53m height , what is the velocity just before it hits the ground? ​

Answer 1

Answer:

Physics of Free-fall Time!!!!!

d = vi * t + 1/2 * g * t^2

499 = 0 * t + 1/2 * 9.8 * t^2

0 * t = 0

499 = 1/2(9.8)(t^2)

2(499) = 9.8 * t^2

(2(499))/9.8 = t^2

998/9.8 = t^2

9.8 = 2 significant figures; 998 = 3 significant figures

2 < 3

We have 2 significant figures.

100 = t^2

100^(1/2) = t

10 seconds = t

And that is the solution plus the proof.

Explanation:

Answer 2

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