Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be y0=0.
Solution:
The given known quantities are:a=-9.8\:\text{m/s}^2; y_o=0\:\text{m}; and v_{oy}=+15\:\text{m/s}.
To compute for the displacement, we use the formula
\Delta y=v_{oy}t+\frac{1}{2}at^2and to compute for the final velocity, we use the formula
v_{fy}=v_{oy}+atPart A
The displacement at t=0.500 \ \text{s} is
\begin{align*}
\Delta y & =v_ot+\frac{1}{2}at^2 \\
\Delta y & =0\:\text{m}+\left(15.0\:\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 & =6.28\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}The velocity at t=0.500 \ \text{s} is
\begin{align*}
v_{fy} & = v_{oy}+at \\
v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(0.500\:\text{s}\right) \\
v_{fy} & =10.1\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Part B
The displacement at t=1.000 \ \text{s} is
\begin{align*}
\Delta y & =v_ot+\frac{1}{2}at^2 \\
\Delta y & =0\:\text{m}+\left(15.0\:\text{m/s}\right)\left(1.000\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.000\:\text{s}\right)^2 \\
\Delta y & =10.1\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}The velocity at t=1.000\ \text{s} is
\begin{align*}
v_{fy} & = v_{oy}+at \\
v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.000\:\text{s}\right) \\
v_{fy} & =5.20\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Part C
The displacement at t=1.500\ \text{s} is
\begin{align*}
\Delta y & =v_ot+\frac{1}{2}at^2 \\
\Delta y & =0\:\text{m}+\left(15.0\:\text{m/s}\right)\left(1.500\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.500\:\text{s}\right)^2 \\
\Delta y & =11.5\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}The velocity at t=1.500\ \text{s} is
\begin{align*}
v_{fy} & = v_{oy}+at \\
v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.500\:\text{s}\right) \\
v_{fy} & =0.300\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Part D
The displacement at t=2.000\ \text{s} is
\begin{align*}
\Delta y & =v_ot+\frac{1}{2}at^2 \\
\Delta y & =0\:\text{m}+\left(15.0\:\text{m/s}\right)\left(2.000\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(2.000\:\text{s}\right)^2 \\
\Delta y & =10.4\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}The velocity at t=2.000\ \text{s} is
\begin{align*}
v_{fy} & = v_{oy}+at \\
v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(2.000\:\text{s}\right) \\
v_{fy} & =-4.600\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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