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A flowerpot falls from a windowsill 15.0 m above the sidewalk.

a. How fast is the flowerpot moving when it strikes the ground?

b. How much time does a passerby on the sidewalk below have to move out of the way before the flowerpot hits the ground?

### 2 Answers

- electron1Lv 71 year ago
As the flowerpot falls 15.0 meters, it has an acceleration of 9.8 m/s^2. Let’s use the following equation to determine its velocity when it strikes the ground.

vf^2 = vi^2 + 2 * a * d, vi = 0

vf^2 = 2 * 9.8 * 15 = 294

vf = √294

This is approximately 17.1 m/s.

b. How much time does a passerby on the sidewalk below have to move out of the way before the flowerpot hits the ground?

Let’s use the following equation to determine the time for this to happen.

vf = vi + a * t, vi = 0

√294 = 9.8 * t

t = √294 ÷ 9.8

This is approximately 1.75 seconds. I hope this is helpful for you.

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- oubaasLv 71 year ago
final speed V = √2gh = √19.612*15 = 17.15 m/sec

falling time t = √2h/g = √30/9.806 = 1.75 sec

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