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If a ball dropped from a tower reaches the ground after 3.5 seconds, what is the height of the tower? Given: g = -9.8 meters/second2 a.) 56 meters b.)60.03 meters c.)62.08 meters d.)62.50 meters
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it is b
\[d=v_0t+\frac{ 1 }{ 2 }at^2\]
\[s=\frac{1}{2}g t^2\] Your answer will come in negative of course due to negative value of g that is taken, it simply means that the displacement is in the same direction as the acceleration, it's magnitude you can take as \[|s|=\frac{1}{2}t^2|g|\]
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