Exercise53

The formula

\begin{equation*} y=\frac{1}{2}\,a\,t^2 +v_0\,t + y_0 \end{equation*}

gives the vertical position of an object, at time $$t\text{,}$$ thrown with an initial velocity $$v_0\text{,}$$ from an initial position $$y_0$$ in a place where the acceleration of gravity is $$a\text{.}$$ The acceleration of gravity on earth is $${-9.8\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s^{2}}}}\text{.}$$ It is negative, because we consider the upward direction as positive in this situation, and gravity pulls down.

What is the height of a baseball thrown with an initial velocity of $$v_0={78\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\text{,}$$ from an initial position of $$y_0= {97\ {\rm m}}\text{,}$$ and at time $$t={14\ {\rm s}}\text{?}$$

Fourteen seconds after the baseball was thrown, it was high in the air.

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