Exercise 54

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 \(v_0={84\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\text{,}\) from an initial position of \(y_0= {80\ {\rm m}}\text{,}\) and at time \(t={5\ {\rm s}}\text{?}\)

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