Vectors And Motion In Two Dimensions

Given: $$F=60\text{ N}, \theta =53^{\circ}$$ Finding the x-component: $${F}_{x}=Fcos\theta $$ $${F}_{x}=60\times cos53^{\circ}$$ $${F}_{x}=60\times 0.6018$$ $${F}_{x}=36.1\text{ N}$$ Finding the y-component: $${F}_{y}=Fsin\theta $$ $${F}_{y}=60\times sin53^{\circ}$$ $${F}_{y}=60\times 0.7986$$ $${F}_{y}=47.9\text{ N}$$ The x-component is 36.1 N and the y-component is 47.9 N.

Given: $${d}_{x}=5.0\text{ m}, {d}_{y}=-12.0\text{ m}$$ Finding the magnitude: $$d=\sqrt{{d}_{x}^{2}+{d}_{y}^{2}}$$ $$d=\sqrt{(5.0{)}^{2}+(-12.0{)}^{2}}$$ $$d=\sqrt{25+144}=\sqrt{169}$$ $$d=13.0\text{ m}$$ Finding the direction: $$\theta ={tan}^{-1}\left(\frac{-12.0}{5.0}\right)={tan}^{-1}(-2.40)$$ $$\theta =-67.4^{\circ}$$ The resultant displacement is 13.0 m directed 67.4° below the positive x-axis. The negative angle indicates the vector points into the fourth quadrant.

Resolve $\vec{P}$: $${P}_{x}=10.0cos30^{\circ}=10.0\times 0.8660=8.66\text{ N}$$ $${P}_{y}=10.0sin30^{\circ}=10.0\times 0.500=5.00\text{ N}$$ Resolve $\vec{Q}$: $${Q}_{x}=8.0cos120^{\circ}=8.0\times (-0.500)=-4.00\text{ N}$$ $${Q}_{y}=8.0sin120^{\circ}=8.0\times 0.8660=6.93\text{ N}$$ Sum components: $${R}_{x}=8.66+(-4.00)=4.66\text{ N}$$ $${R}_{y}=5.00+6.93=11.93\text{ N}$$ Finding the magnitude: $$R=\sqrt{(4.66{)}^{2}+(11.93{)}^{2}}$$ $$R=\sqrt{21.7+142.3}=\sqrt{164.0}$$ $$R=12.8\text{ N}$$ The resultant force has a magnitude of 12.8 N.

The resultant can be zero. If two vectors have equal magnitude and point in exactly opposite directions, their x-components cancel and their y-components cancel, so ${R}_{x}=0$ and ${R}_{y}=0$. The resultant magnitude is therefore zero.

The horizontal and vertical components of motion are perpendicular. Because the acceleration (gravity) acts only along the vertical axis, it changes the vertical velocity but has no effect on the horizontal velocity. Each axis therefore satisfies its own kinematic equations, linked only through the shared time variable.

Finding ${v}_{x}$ at $t=2.0$ s: $${v}_{x}={v}_{0x}+{a}_{x}t=4.0+(0)(2.0)=4.0\text{ m/s}$$ Finding ${v}_{y}$ at $t=2.0$ s: $${v}_{y}={v}_{0y}+{a}_{y}t=3.0+(-2.0)(2.0)=3.0-4.0=-1.0\text{ m/s}$$ Finding the speed: $$v=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}$$ $$v=\sqrt{(4.0{)}^{2}+(-1.0{)}^{2}}=\sqrt{16.0+1.0}=\sqrt{17.0}$$ $$v=4.12\text{ m/s}$$ The speed at $t=2.0$ s is 4.12 m/s.

Because the launch is horizontal, ${v}_{0y}=0$. Given: $$\Delta y=-80.0\text{ m}, {v}_{0y}=0, g=10{\text{ m/s}}^{2}$$ Finding the time to fall: $$\Delta y={v}_{0y}t-\frac{1}{2}g{t}^{2}$$ $$-80.0=0-\frac{1}{2}(10){t}^{2}$$ $$-80.0=-5.00 {t}^{2}$$ $${t}^{2}=\frac{80.0}{5.00}=16.0$$ $$t=4.00\text{ s}$$ Finding the horizontal distance: $$\Delta x={v}_{0x}\times t=15.0\times 4.00=60.0\text{ m}$$ The projectile hits the ground after 4.00 s, having traveled 60.0 m horizontally.

Doubling the launch speed doubles the time of flight. The time of flight is $t=\frac{2{v}_{0}sin\theta }{g}$, which is directly proportional to ${v}_{0}$. Because $sin\theta $ and $g$ remain unchanged, doubling ${v}_{0}$ doubles $t$.

Starting point: Use the kinematic equation relating velocity, initial velocity, acceleration, and displacement. $${v}_{y}^{2}={v}_{0y}^{2}+2{a}_{y}\Delta y$$ Step 1 — Identify conditions at maximum height At the peak, the vertical velocity is zero, so ${v}_{y}=0$. The vertical acceleration is ${a}_{y}=-g$. The vertical displacement is $\Delta y=H$. $$0={v}_{0y}^{2}+2(-g)(H)$$ Step 2 — Substitute the vertical component of initial velocity $${v}_{0y}={v}_{0}sin\theta $$ $$0=({v}_{0}sin\theta {)}^{2}-2gH$$ Step 3 — Solve for $H$ $$2gH={v}_{0}^{2}{sin}^{2}\theta $$ $$H=\frac{{v}_{0}^{2}{sin}^{2}\theta }{2g}$$ Result: $$H=\frac{{v}_{0}^{2}{sin}^{2}\theta }{2g}$$ Maximum height is directly proportional to the square of the launch speed. Doubling ${v}_{0}$ quadruples $H$.

The expression is consistent. Substituting $\theta =90^{\circ}$ gives ${sin}^{2}90^{\circ}=1$, so $H=\frac{{v}_{0}^{2}}{2g}$, which is the largest possible value of $H$ because ${sin}^{2}\theta \le 1$ for all angles. Any other angle produces a smaller ${sin}^{2}\theta $ and therefore a smaller maximum height.