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ddd [48]
3 years ago
5

While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 6.95 6.95 m/s. The st

one subsequently falls to the ground, which is 14.3 14.3 m below the point where the stone leaves his hand. At what speed does the stone impact the ground? Ignore air resistance and use g = 9.81 m/s 2 g=9.81 m/s2 for the acceleration due to gravity.
Physics
1 answer:
qwelly [4]3 years ago
8 0

Answer:

18.1347 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-6.95^2}{2\times -9.81}\\\Rightarrow s=2.4619\ m

Total height the ball falls is 2.4619+14.3 = 16.7619 m

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 16.7619+0^2}\\\Rightarrow v=18.1347\ m/s

The speed at which the stone reaches the ground is 18.1347 m/s

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2 years ago
The 6 strings on a guitar all have about the same length and are stretched with about the same tension.The highest string vibrat
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f = (1/2L)√(T/((p(πd^2)L/4)/L))

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Since the length of the strings are the same, the frequency is inversely proportional to the string diameter.

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Answer:

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