Using
V = Amplitude x angular frequency(omega)
But omega= 2πf
= 2πx875
=5498.5rad/s
So v= 1.25mm x 5498.5
= 6.82m/s
B. .Acceleration is omega² x radius= 104ms²
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SOOOOO THE ANSWER IS ACRONYMIC SENTENCE IM 99.99% SURE</em></h3>
Answer:
12.5m/s
Explanation:
(Assuming the question was asking for the speed just before it hit the ground)
We can use the first key equation of accelerated motion
Vf^2 = Vi^2+2aΔd
Vf^2 = 0 + 2(9.8)(8) (plugged in values, initial velocity is 0 since the ball was at rest)
Vf^2 = 156.8
Vf = 12.5 (squared both sides)
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