Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.
Solution:
The given known quantities are: a=-9.8\:\text{m/s}^2; y_0=0 \ \text{m}; and v_0=-14 \ \text{m/s}.
To compute for the displacement, we use the formula
\Delta y=v_0t+\frac{1}{2}at^2and to compute for the final velocity, we use the formula
v_f=v_0+at
Part A
The displacement at t=0.500 \ \text{s} is
\begin{align*}
\Delta y & =v_0t+\frac{1}{2}at^2 \\
\Delta y &=\left(-14.0\:\text{m/s}\right)\left(0.500\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(0.500\:\text{s}\right)^2 \\
\Delta y & =-8.23\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
The velocity at t=0.500 \ \text{s} is
\begin{align*}
v_f & =v_0+at \\
&= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(0.500\:\text{s}\right) \\
& =-18.9\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
Part B
The displacement at t=1.00\ \text{s} is
\begin{align*}
\Delta y & =v_0t+\frac{1}{2}at^2 \\
\Delta y &=\left(-14.0\:\text{m/s}\right)\left(1.00\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.00\:\text{s}\right)^2 \\
\Delta y & =-18.9\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
The velocity at t=1.00\ \text{s} is
\begin{align*}
v_f & =v_0+at \\
&= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.00\:\text{s}\right) \\
& =-23.8\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
Part C
The displacement at t=1.50\ \text{s} is
\begin{align*}
\Delta y & =v_0t+\frac{1}{2}at^2 \\
\Delta y &=\left(-14.0\:\text{m/s}\right)\left(1.50\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.50\:\text{s}\right)^2 \\
\Delta y & =-32.0\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
The velocity at t=1.50\ \text{s} is
\begin{align*}
v_f & =v_0+at \\
&= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.50\:\text{s}\right) \\
& =-28.7\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
Part D
The displacement at t=2.00\ \text{s} is
\begin{align*}
\Delta y & =v_0t+\frac{1}{2}at^2 \\
\Delta y &=\left(-14.0\:\text{m/s}\right)\left(2.00\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(2.00\:\text{s}\right)^2 \\
\Delta y & =-47.6\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
The velocity at t= 2.00 \ \text{s} is
\begin{align*}
v_f & =v_0+at \\
&= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(2.00\:\text{s}\right) \\
& =-33.6\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
Part E
The displacement at t=2.50\ \text{s} is
\begin{align*}
\Delta y & =v_0t+\frac{1}{2}at^2 \\
\Delta y &=\left(-14.0\:\text{m/s}\right)\left(2.50\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(2.50\:\text{s}\right)^2 \\
\Delta y & =-65.6\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
The velocity at t= 2.50 \ \text{s} is
\begin{align*}
v_f & =v_0+at \\
&= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(2.50\:\text{s}\right) \\
& =-38.5\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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