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Your friend is standing below the window outside of our classroom and attempting to throw a rock past the window to get your attention. The rock is thrown straight up in the air and passes our window 0.40s after being released. It passes the same window on its way back down 1.60s later. What was the initial velocity of the rock? Did the height of the window matter?

Nick Peterson
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Sandro Hallado
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    Did you try drawing a picture? Do you know any equations that relate initial velocity and final velocity? Have you written all the information given and tried organizing it? – Amzoti Sep 03 '13 at 04:14
  • Depends on the meaning of passes. If first pass is when the rock just becomes visible, and second pass is when rock ceases to be visible on the way down, then height of the window obviously doesn't matter. Use the relationship $s=v_0 t-\frac{1}{2}gt^2$. – André Nicolas Sep 03 '13 at 04:23
  • Might the question be the height of the window above the ground, not the height from top to bottom? – DJohnM Sep 03 '13 at 05:12

3 Answers3

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Don't know if the question is on-topic for https://math.stackexchange.com/ but the approach will be as follows:

Let initial velocity be $ \vec{u} $ towards the positive y-direction (up). Acceleration due to gravity is $ - \vec{g} $ towards negative y-direction (down).

Use Newton's equations of linear motion:

$$ \begin{align} \vec{v} &= \vec{u} + t * \vec{a} \\ \vec{s} &= t * \vec{u} + \frac{1}{2} t^2 * \vec{a} \\ v^2 &= u^2 + 2 * a * s \end{align} $$

You'll only need the second equation for two different values of $ t $.

Community
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hjpotter92
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\begin{align} \vec{y}_{f} &= \vec{y}_{0} + \vec{v}t-\frac{1}{2}gt^2 \end{align} Where \begin{align}\vec{y}_0\end{align} is the initial height and yf is the final height.

pyler
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The rock took $0.4$ seconds to reach the window, and a further $1.6$ seconds to pass the window, come to a stop, turn around and come back down to the window. So the rock was going up for a total of $0.4+1.6/2$. or $1.2$ seconds. In that time it lost $1.2\times9.8$, or $11.8$ m/s of velocity ($g=9.8 \frac{m}{s^2}$). So that must be the initial velocity.

DJohnM
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