Initial Velocity of a Thrown Rock

Question

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?

Answer 1

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:

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

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

Answer 2

\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.

Answer 3

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.