Answer Key Chapter 3 - College Physics for AP® Courses | OpenStax (2024)

(a) $\text{480 m}$

(b) $\text{379 m}$, $\text{18.4°}$ east of north

(a) $\text{26}\text{.}\text{6 m}$, $\text{65}\text{.}\text{1°}$ north of east

(b) $\text{26}\text{.}\text{6 m}$, $\text{65}\text{.}\text{1°}$ south of west

x-component 4.41 m/s

y-component 5.07 m/s

(a) 1.56 km

(b) 120 m east

(a) $\text{30}\text{.}\text{8 m}$, $\text{54}\text{.}\mathrm{2º}$ south of west

(b) $\text{30}\text{.}\text{8 m}$, $\text{54}\text{.}\mathrm{2º}$ north of east

(a) 3.50 s

(b) 28.6 m/s (c) 34.3 m/s

(d) 44.7 m/s, $\mathrm{50.2°}$ below horizontal

(a) $\text{18}\text{.}\text{4°}$

(b) The arrow will go over the branch.

$\begin{array}{}R=\frac{v_0^2}{\text{sin}{\mathrm{2\theta }}_{0}g}\\ \text{For}\theta =\text{45°},R=\frac{v_0^2}{g}\end{array}$

$R=91.8\text{m}$ for$v_0=30\text{m/s}$;$R=163\text{m}$for$v_0=40\text{m/s}$;$R=255\text{m}$ for$v_0=50\text{m/s}$.

(a) 560 m/s

(b) $8\text{.}\text{00}\times {\text{10}}^3\text{m}$

(c) 80.0 m. This error is not significant because it is only 1% of the answer in part (b).

$\theta =\mathrm{6.1°}$

yes, the ball lands at 5.3 m from the net

(a) −0.486 m

(b) The larger the muzzle velocity, the smaller the deviation in the vertical direction, because the time of flight would be smaller. Air resistance would have the effect of decreasing the time of flight, therefore increasing the vertical deviation.

(a) 24.2 m/s

(b) The ball travels a total of 57.4 m with the brief gust of wind.

$y-y_0=0=v_{0y}t-\frac{1}{2}{\mathrm{gt}}^2=(v_0\text{sin}\theta )t-\frac{1}{2}{\mathrm{gt}}^2$,

so that $t=\frac{2(v_0\text{sin}\theta )}{g}$

$x-x_0=v_{0x}t=(v_0\text{cos}\theta )t=R,$ and substituting for $t$ gives:

$R=v_0\text{cos}\theta \left(\frac{{2v}_0\text{sin}\theta }{g}\right)=\frac{{2v}_0^2\text{sin}\theta \text{cos}\theta }{g}$

since $2\text{sin}\theta \text{cos}\theta =\text{sin}\mathrm{2\theta },$ the range is:

$R=\frac{{v_0}^2\text{sin}\mathrm{2\theta }}{g}$.

(a) $\text{35}\text{.}\mathrm{8\; km}$, $\text{45º}$ south of east

(b) $5\text{.}\text{53 m/s}$, $\text{45º}$ south of east

(c) $\text{56}\text{.}\mathrm{1\; km}$, $\text{45º}$ south of east

(a) 0.70 m/s faster

(b) Second runner wins

(c) 4.17 m

(a) $2\text{30 m/s}$, $\mathrm{8.0º}$ south of west

(b) The wind should make the plane travel slower and more to the south, which is what was calculated.

(a) 63.5 m/s

(b) 29.6 m/s

(a) $H_{\text{average}}=\text{14}\text{.}\text{9}\frac{\text{km/s}}{\text{Mly}}$

(b) 20.2 billion years

The graph of the ball's vertical velocity over time should begin at 4.90 m/s during the time interval 0 - 0.1 sec (there should be a data point at t = 0.05 sec, v = 4.90 m/s). It should then have a slope of -9.8 m/s², crossing through v = 0 at t = 0.55 sec and ending at v = -0.98 m/s at t = 0.65 sec.

The graph of the ball's horizontal velocity would be a constant positive value, a flat horizontal line at some positive velocity from t = 0 until t = 0.7 sec.