A stone dropped from the root of a high buliding. A second stone is dropped 1.30 s later How far apart are the stones when the second one has reached a speed of 14.0 m/s

Question

Physics

Mitzi

8 March, 16:33

A stone dropped from the root of a high buliding. A second stone is dropped 1.30 s later How far apart are the stones when the second one has reached a speed of 14.0 m/s

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Answers (1)

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Drew Castro · Answer 1

The stones are apart one each the other 26.37 meters when the second one has reached a speed of 14 m/s.

Explanation:

Second Stone:

Vb = g*t

tb (14m/s) = Vb/g

tb (14m/s) = (14 m/s) / (9.8 m/s²)

tb (14m/s) = 1.42 sec

Hb = (g * tb²) / 2

Hb = 9.88m

First Stone:

ta = tb + 1.3 sec

ta = 2.72 sec

Ha = (g*ta²) / 2

Ha = 36.25m

Distance between Stones:

Ha-Hb = 36.25m - 9.88m

Ha-Hb = 26.37m