College Physics by Openstax Chapter 2 Problem 41: Displacement and Velocity of a Vertically Thrown Ball at Successive Time Intervals

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^2

and to compute for the final velocity, we use the formula

v_{fy}=v_{oy}+at

Part 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*}