(a) $\text{480 m}$

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

north component 3.21 km, east component 3.83 km

$\text{19}\text{.}\text{5 m}$, $4\text{.}\text{65º}$ south of west

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

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

$\text{52}\text{.}\text{9 m}$, $\text{90}\text{.}\text{1º}$ with respect to the x-axis.

x-component 4.41 m/s

y-component 5.07 m/s

(a) 1.56 km

(b) 120 m east

North-component 87.0 km, east-component 87.0 km

30.8 m, 35.8 west of north

(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

18.4 km south, then 26.2 km west(b) 31.5 km at $\mathrm{45.0º}$ south of west, then 5.56 km at $\mathrm{45.0º}$ west of north

$7\text{.}\text{34 km}$, $\text{63}\text{.}\mathrm{5º}$ south of east

$\begin{array}{lll}x& =& \text{1.30 m}\times {10}^2\\ y& =& \text{30}\text{.9 m.}\end{array}$

(a) 3.50 s

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

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

(a) $\text{18}\text{.}\text{4º}$

(b) The arrow will go over the branch.

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

$R=91.8\text{m}$ 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).

1.50 m, assuming launch angle of $\mathrm{45º}$

$\theta =\mathrm{6.1º}$

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.

4.23 m. No, the owl is not lucky; he misses the nest.

No, the maximum range (neglecting air resistance) is about 92 m.

15.0 m/s

(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

$17\text{.}\text{0 m/s}$, $\text{22}\text{.}\mathrm{1º}$

(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

$6\text{.}\text{68 m/s}$, $\text{53}\text{.}\mathrm{3º}$ south of west

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

(b) 20.2 billion years

$1\text{.}\text{72 m/s}$, $\mathrm{42.3º}$ north of east

(a) Since ${\sin }^{2}50>{\sin }^{2}40$, then B reaches the greatest height.

(b) i. Yes, it is consistent because $\sin 50>\sin 40$. ii. No, it does not make sense because y is proportional to ${\sin }^2\theta$.

(c)

Figure 3.61

(d)

We would need to know the horizontal and vertical positions of each ball at several times. From that data, we could deduce the velocities over several time intervals and also the accelerations (both horizontal and vertical) for each ball over several time intervals.

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.