College Physics by Openstax Chapter 2 Problem 41

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 m/s2a=-9.8\:\text{m/s}^2; yo=0 my_o=0\:\text{m}; and voy=+15 m/sv_{oy}=+15\:\text{m/s}.

To compute for the displacement, we use the formula

Δy=voyt+12at2\Delta y=v_{oy}t+\frac{1}{2}at^2

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

vfy=voy+atv_{fy}=v_{oy}+at

Part A

The displacement at t=0.500 st=0.500 \ \text{s} is

Δy=vot+12at2Δy=0 m+(15.0 m/s)(0.500 s)+12(−9.8 m/s2)(0.500 s)2Δy=6.28 m  (Answer)\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 st=0.500 \ \text{s} is

vfy=voy+atvfy=(15.0 m/s)+(−9.8 m/s2)(0.500 s)vfy=10.1 m/s  (Answer)\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 st=1.000 \ \text{s} is

Δy=vot+12at2Δy=0 m+(15.0 m/s)(1.000 s)+12(−9.8 m/s2)(1.000 s)2Δy=10.1 m  (Answer)\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 st=1.000\ \text{s} is

vfy=voy+atvfy=(15.0 m/s)+(−9.8 m/s2)(1.000 s)vfy=5.20 m/s  (Answer)\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 st=1.500\ \text{s} is

Δy=vot+12at2Δy=0 m+(15.0 m/s)(1.500 s)+12(−9.8 m/s2)(1.500 s)2Δy=11.5 m  (Answer)\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 st=1.500\ \text{s} is

vfy=voy+atvfy=(15.0 m/s)+(−9.8 m/s2)(1.500 s)vfy=0.300 m/s  (Answer)\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 st=2.000\ \text{s} is

Δy=vot+12at2Δy=0 m+(15.0 m/s)(2.000 s)+12(−9.8 m/s2)(2.000 s)2Δy=10.4 m  (Answer)\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 st=2.000\ \text{s} is

vfy=voy+atvfy=(15.0 m/s)+(−9.8 m/s2)(2.000 s)vfy=−4.600 m/s  (Answer)\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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