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An object drops from a building and strikes the ground with a speed of 31m/s.How tall is the building?

Respuesta :

Lanuel

Answer:

49.0m

Explanation:

Given the following data;

U = 0m/s since it's a free fall

V = 31m/s

Also, since it's a free fall, a = g = 9.8m/s²

To find the height of the building, we would use the third equation of motion;

[tex] V^{2} = U^{2} + 2aS [/tex]

Where;

  • V represents the final velocity measured in meter per seconds.
  • U represents the initial velocity measured in meter per seconds.
  • a represents acceleration due to gravity measured in meters per seconds square.
  • S represents the height measured in meters.

Substituting the values into the equation, we have;

[tex] 31^{2} = 0^{2} + 2*9.8*S [/tex]

[tex] 961 = 0 + 19.6S [/tex]

Simplifying, we have;

[tex] S = \frac {961}{19.6}[/tex]

S = 49.03 ≈ 49.0

Therefore, the building is 49.0m tall.

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