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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Fletcher Madden

06.10.20, 11:25

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

Answer to Question #135408 in Mechanics | Relativity for Vanel

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