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Physics 22 Online
OpenStudy (anonymous):

a stone is thrown vertically at 20/s upward. how long does it take to reach the peak of its motion, and how high does it go?

OpenStudy (ybarrap):

Is the speed initial speed 20 m/s ?

OpenStudy (anonymous):

yes it is

OpenStudy (ybarrap):

$$ \begin{align}v & = u + at \quad [1] \\s & = ut + \frac{1}{2} at^2 \quad [2] \\s & = \frac{1}{2}(u + v)t \quad [3] \\v^2 & = u^2 + 2as \quad [4] \\s & = vt - \frac{1}{2}at^2 \quad [5] \\\end{align} $$ Using [1] above, you can solve for \(t\), the time to reach the peak. Take \(v=0,u=20\) and \(a=-9.81\). This is because the final velocity at the peak is zero. You can then use this \(t\) in [2],[3] or [5] to find out how high the stone gets.

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