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Answer to Question #135408 in Mechanics | Relativity for Vanel
Question #135408
A rock is thrown downward from the top of a 40.0m tall tower with an initial speed of 12 m/s. Assuming negligible air resistance, what is the speed of the rock just before hitting the ground?
Expert's answer
"h=\\frac{V^2-V_0^2}{2g};\\\\V=\\sqrt{2gh+V_0}=\\sqrt{2\\times9.8\\times40+12}=28.21\\frac{m}{s};\\\\Answer: the \\;speed\\ of\\ the\\ rock\\ is \\;28.21\\frac{m}{s};"
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Comments
Hello dear experts! There's a minute error in the substitution of values as; v^2=vo^2+2gh v=sqrt(2gh+vo^2) v=sqrt[2x9.8x40 + (12)^2] = 30.46m/s Hope you guys figure out this correction and thanks for your endervous help
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