{"id":1009,"date":"2025-10-15T13:01:59","date_gmt":"2025-10-15T13:01:59","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1009"},"modified":"2025-10-23T12:28:22","modified_gmt":"2025-10-23T12:28:22","slug":"motion-in-space-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/motion-in-space-learn-it-3\/","title":{"raw":"Motion in Space: Learn It 3","rendered":"Motion in Space: Learn It 3"},"content":{"raw":"

Projectile Motion<\/h2>\r\n

Now let's apply vector functions to a real-world scenario: understanding how gravity affects objects moving through the air. This situation\u2014where an object moves with an initial velocity but experiences only the force of gravity\u2014is called projectile motion<\/strong>. It describes everything from golf balls and baseballs to arrows and cannonballs. We'll ignore air resistance in our analysis.<\/p>\r\n\r\n

Setting Up the Coordinate System<\/h3>\r\n

We place ourselves at the origin with the positive [latex]y[\/latex]-axis pointing up, the negative [latex]y[\/latex]-axis pointing down, and the positive [latex]x[\/latex]-axis pointing forward (away from the thrower). Since gravity acts downward, Newton's second law tells us the force from gravity equals the mass times the acceleration due to gravity: [latex]F_{g}=mg[\/latex], where [latex]F_{g}[\/latex] is the gravitational force and [latex]g[\/latex] is the acceleration due to gravity at Earth's surface.<\/p>\r\n

The value of [latex]g[\/latex] is approximately [latex]32 ft\/sec^{2}[\/latex] in the English system and approximately [latex]9.8 m\/sec^{2}[\/latex] in the metric system. Since gravity acts downward, we write the gravitational force as [latex]F_{g}=-mg{\\bf{j}}[\/latex], as shown in Figure 1.<\/p>\r\n\r\n\r\n[caption id=\"attachment_929\" align=\"aligncenter\" width=\"439\"]\"This Figure 1. An object is falling under the influence of gravity.[\/caption]\r\n

Deriving the Position Equation<\/h3>\r\n

Newton's second law states that [latex]F=m{\\bf{a}}[\/latex], where [latex]{\\bf{a}}[\/latex] is the acceleration vector. This force must equal the gravitational force at all times:<\/p>\r\n

[latex]\\begin{array}{ccc}\\hfill {F} & =\\hfill & {F_{g}} \\hfill \\\\ \\hfill {m{\\bf{a}}}& =\\hfill & {-mg{\\bf{j}}} \\hfill \\\\ \\hfill {{\\bf{a}}}& =\\hfill & {-g{\\bf{j}}}\\hfill \\end{array}[\/latex]<\/p>\r\n

Since the acceleration vector is the derivative of the velocity vector, we can write:<\/p>\r\n

[latex]{{\\bf{v}}'(t)=-g{\\bf{j}}}[\/latex]<\/p>\r\n

Taking the antiderivative of both sides:<\/p>\r\n

[latex]\\begin{array}{ccc}\\hfill {{\\bf{v}}(t)} & =\\hfill & {\\int -g{\\bf{j}}dt} \\hfill \\\\ \\hfill & =\\hfill & {-gt{\\bf{j}}+{\\bf{C}}_{1}}\\hfill \\end{array}[\/latex]<\/p>\r\n

To find the constant vector [latex]{\\bf{C}}{1}[\/latex], we use the initial velocity<\/strong> [latex]{\\bf{v}}{0}={\\bf{v}}(0)[\/latex]. This gives us [latex]{\\bf{C}}{1}={\\bf{v}}{0}[\/latex], so:<\/p>\r\n

[latex]{\\bf{v}}(t)=-gt{\\bf{j}}+{\\bf{v}}_{0}[\/latex]<\/p>\r\n

Since velocity is the derivative of position, we have:<\/p>\r\n

[latex]{\\bf{s}}'(t)=-gt{\\bf{j}}+{\\bf{v}}_{0}[\/latex]<\/p>\r\n

Taking the antiderivative again:<\/p>\r\n

[latex]\\begin{array}{ccc}\\hfill {{\\bf{s}}(t)} & =\\hfill & {\\int -gt{\\bf{j}}+{\\bf{v}}{0}dt} \\hfill \\\\ \\hfill & =\\hfill & {-\\frac{1}{2}gt^{2}{\\bf{j}}+{\\bf{v}}{0}t+{\\bf{C}}_{2}}\\hfill \\end{array}[\/latex]<\/p>\r\n

Using the initial position<\/strong> [latex]{\\bf{s}}{0}={\\bf{s}}(0)[\/latex], we find [latex]{\\bf{C}}{2}={\\bf{s}}_{0}[\/latex]. This gives the position at any time:<\/p>\r\n

[latex]{\\bf{s}}(t)=-\\frac{1}{2}gt^{2}{\\bf{j}}+{\\bf{v}}{0}t+{\\bf{s}}{0}[\/latex]<\/p>\r\n\r\n

Projectile Launched at an Angle<\/h3>\r\n

Suppose an object is thrown upward from the origin at an angle [latex]\\theta[\/latex] to the horizontal with initial speed [latex]v_{0}[\/latex]. We can assume [latex]{\\bf{s}}_{0}={\\bf{0}}[\/latex] by placing the origin at the launch point (Figure 2).<\/p>\r\n\r\n[caption id=\"attachment_930\" align=\"aligncenter\" width=\"650\"]\"This Figure 2. Projectile motion when the object is thrown upward at an angle [latex]\\theta[\/latex]. The horizontal motion is at constant velocity and the vertical motion is at constant acceleration.[\/caption]\r\n

We can express the initial velocity as [latex]{\\bf{v}}{0}=v{0}\\cos\\theta{\\bf{i}}+v_{0}\\sin\\theta{\\bf{j}}[\/latex]. The position equation becomes:<\/p>\r\n

[latex]\\begin{array}{ccc}\\hfill {{\\bf{s}}(t)} & =\\hfill & {-\\frac{1}{2}gt^{2}{\\bf{j}}+v_{0}t\\cos\\theta{\\bf{i}}+v_{0}t\\sin\\theta{\\bf{j}}} \\hfill \\\\ \\hfill & =\\hfill & {v_{0}t\\cos\\theta{\\bf{i}}+(v_{0}t\\sin\\theta-\\frac{1}{2}gt^{2}){\\bf{j}}}\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n

\r\n

components of projectile motion<\/h3>\r\n
    \r\n \t
  • The coefficient of [latex]{\\bf{i}}[\/latex] represents the horizontal distance<\/strong> from the origin at time [latex]t[\/latex]<\/li>\r\n \t
  • The maximum horizontal distance (at the same initial and final altitude) is called the range<\/strong> [latex]R[\/latex]<\/li>\r\n \t
  • The coefficient of [latex]{\\bf{j}}[\/latex] represents the altitude<\/strong> at time [latex]t[\/latex]<\/li>\r\n \t
  • The maximum altitude is called the height<\/strong> [latex]H[\/latex]<\/li>\r\n<\/ul>\r\n<\/section>
    During an Independence Day celebration, a cannonball is fired from a cannon on a cliff toward the water. The cannon is aimed at an angle of [latex]30^\\circ[\/latex] above horizontal and the initial speed of the cannonball is [latex]600\\ ft\/sec[\/latex]. The cliff is [latex]100\\ ft[\/latex] above the water (Figure 7).\r\n
      \r\n \t
    1. Find the maximum height of the cannonball.<\/li>\r\n \t
    2. How long will it take for the cannonball to splash into the sea?<\/li>\r\n \t
    3. How far out to sea will the cannonball hit the water?\r\n\r\n[caption id=\"attachment_931\" align=\"aligncenter\" width=\"692\"]\"This Figure 3. The flight of a cannonball (ignoring air resistance) is projectile motion.[\/caption]<\/li>\r\n<\/ol>\r\n
      [reveal-answer q=\"fs-id1267795055160\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1267795055160\"]We use the equation\r\n
      [latex]{\\bf{s}}(t)=v_{0}t\\cos{\\theta{\\bf{i}}}+\\big(v_{0}t\\sin{\\theta}-\\frac{1}{2}gt^{2}\\big){\\bf{j}}[\/latex]<\/div>\r\n
      <\/div>\r\nwith [latex]\\theta={30^\\circ}[\/latex], [latex]g=32\\ ft\/sec^{2}[\/latex], and [latex]v_{0}=600\\ ft\/sec[\/latex]. Then the position equation becomes<\/span>\r\n

      [latex]\\begin{array}{ccc}\\hfill {{\\bf{s}}(t)} & =\\hfill & {600t(\\cos{30}){\\bf{i}}+(600t\\sin{30}-\\frac{1}{2}(32)t^{2}){\\bf{j}}} \\hfill \\\\ \\hfill & =\\hfill & {300t\\sqrt{3}{\\bf{i}}+(300t-16t^{2}){\\bf{j}}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\r\n\r\n

        \r\n \t
      1. The cannonball reaches its maximum height when the vertical component of its velocity is zero, because the cannonball is neither rising nor falling at that point. The velocity vector is\r\n
        <\/div>\r\n
        [latex]\\begin{array}{ccc}\\hfill {{\\bf{v}}(t)} & =\\hfill & {{\\bf{s}}'(t)} \\hfill \\\\ \\hfill & =\\hfill & {300\\sqrt{3}{\\bf{i}}+(300-32t){\\bf{j}}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\nTherefore, the vertical component of velocity is given by the expression [latex]300-32t[\/latex]. Setting this expression equal to zero and solving [latex]t[\/latex] gives [latex]t=9.375\\ sec[\/latex]. The height of the cannonball at this time is given by the vertical component of the position vector, evaluated at [latex]t=9.375[\/latex].\r\n
        [latex]\\begin{array}{ccc}\\hfill {{\\bf{s}}(9.375)} & =\\hfill & {300(9.375)\\sqrt{3}{\\bf{i}}+(300(9.375)-16(9.375)^{2}){\\bf{j}}} \\hfill \\\\ \\hfill & =\\hfill & {4871.39{\\bf{i}}+1406.25{\\bf{j}}}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\nTherefore, the maximum height of the cannonball is [latex]1406.39\\text{ }ft[\/latex] above the cannon, or [latex]1506.39\\text{ }ft[\/latex] above sea level.<\/li>\r\n \t
      2. When the cannonball lands in the water, is is [latex]100\\text{ }ft[\/latex] below the cannon. Therefore, the vertical component of the position vector is equal to [latex]-100[\/latex]. Setting the vertical component of [latex]{\\bf{s}}(t)[\/latex] equal to [latex]-100[\/latex] and solving, we obtain\r\n
        [latex]\\begin{array}{ccc}\\hfill {300t-16t^{2}} & =\\hfill & {-100} \\hfill \\\\ \\hfill {16t^{2}-300t-100}& =\\hfill & {0} \\hfill \\\\ \\hfill {4t^{2}-75t-25}& =\\hfill & {0} \\hfill \\\\ \\hfill {t}& =\\hfill & {\\frac{75\\pm\\sqrt{(-75)^{2}-4(4)(-25)}}{2(4)}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{75\\pm\\sqrt{6025}}{8}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{75\\pm5\\sqrt{241}}{8}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\nThe positive value of [latex]t[\/latex] that solves this equation is approximately [latex]19.08[\/latex]. Therefore, the cannonball hits the water after approximately [latex]19.08\\text{ }sec[\/latex].<\/li>\r\n \t
      3. To find the distance out to sea, we simply substitute the answer from part (b) into [latex]{\\bf{s}}(t)[\/latex]:\r\n
        [latex]\\begin{array}{ccc}\\hfill {{\\bf{s}}(19.08)} & =\\hfill & {300(19.08)\\sqrt{3}{\\bf{i}}+(300(19.08)-16(19.08)^{2}){\\bf{j}}} \\hfill \\\\ \\hfill & =\\hfill & {9914.26{\\bf{i}}-100.7424{\\bf{j}}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\nTherefore, the ball hits the water about [latex]9914.26\\text{ }ft[\/latex] away from the base of the cliff. Notice that the vertical component of the position vector is very close to [latex]-100[\/latex], which tells us that the ball just hit the water. Note that [latex]9914.26[\/latex] feet is not the true range of the cannon since the cannonball lands in the ocean at a location below the cannon. The range of the cannon would be determined by finding how far out the cannonball is when its height is [latex]100\\text{ }ft[\/latex] above the water (the same as the altitude of the cannon).<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

        Maximum Range of a Projectile<\/h3>\r\n

        One final question: what is the maximum distance a projectile can travel given its initial speed? To answer this, we assume the projectile is fired from ground level and returns to ground level. The equation of projectile motion is:<\/p>\r\n

        [latex]{{\\bf{s}}(t)=v_{0}t\\cos\\theta{\\bf{i}}+(v_{0}t\\sin\\theta-\\frac{1}{2}gt^{2}){\\bf{j}}}[\/latex]<\/p>\r\n

        Setting the vertical component equal to zero and solving for [latex]t[\/latex]:<\/p>\r\n

        [latex]\\begin{array}{ccc}\\hfill {v_{0}t\\sin\\theta-\\frac{1}{2}gt^{2}} & =\\hfill & {0} \\hfill \\\\ \\hfill {t(v_{0}\\sin\\theta-\\frac{1}{2}gt)}& =\\hfill & {0}\\hfill \\end{array}[\/latex]<\/p>\r\n

        This gives [latex]t=0[\/latex] or [latex]t=\\frac{2v_{0}\\sin\\theta}{g}[\/latex]. Substituting the second value into [latex]{\\bf{s}}(t)[\/latex]:<\/p>\r\n

        [latex]\\begin{array}{ccc}\\hfill{{\\bf{s}}\\big(\\frac{2v_{0}\\sin{\\theta}}{g}\\big)} & =\\hfill & {v_{0}\\big(\\frac{2v_{0}\\sin{\\theta}}{g}\\big)\\cos{\\theta{\\bf{i}}}+\\Big(v_{0}\\big(\\frac{2v_{0}\\sin{\\theta}}{g}\\big)\\sin{\\theta}-\\frac{1}{2}g\\big(\\frac{2v_{0}\\sin{\\theta}}{g}\\big)^{2}\\Big){\\bf{j}}} \\hfill \\\\ \\hfill & =\\hfill & {\\bigg(\\frac{2v^2_0\\sin{\\theta}\\cos{\\theta}}{g}\\bigg){\\bf{i}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{v^2_0\\sin{2\\theta}}{g}{\\bf{i}}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\r\nThus, the expression for the range of a projectile fired at an angle [latex]\\theta[\/latex] is\r\n

        [latex]\\large{R=\\frac{v^2_0\\sin{2\\theta}}{g}{\\bf{i}}}. [\/latex]<\/p>\r\n\r\n

        \r\n

        range of a projectile<\/h3>\r\n

        The range of a projectile fired at angle [latex]\\theta[\/latex] with initial speed [latex]v_{0}[\/latex] is:<\/p>\r\n

        [latex]R=\\frac{v^2_0\\sin 2\\theta}{g}[\/latex]<\/p>\r\n\r\n<\/section>\r\n

        To maximize the range, take the derivative with respect to [latex]\\theta[\/latex] and set it equal to zero:<\/p>\r\n

        [latex]\\begin{align} \\frac{d}{d\\theta}\\left(\\frac{v_{0}^2\\sin 2\\theta}{g}\\right) & =0 \\\\ \\frac{2v_{0}^2\\cos 2\\theta}{g} & = 0 \\\\ \\theta & = 45^\\circ \\end{align}[\/latex]<\/p>\r\n\r\n

        \r\n

        In the absence of air resistance, the angle that maximizes range is [latex]45^\\circ[\/latex]. At this angle, the range is:<\/p>\r\n

        [latex]R=\\frac{v^2_0}{g}[\/latex]<\/p>\r\n

        This result assumes the projectile is launched and lands at the same altitude.<\/p>\r\n\r\n<\/section>","rendered":"

        Projectile Motion<\/h2>\n

        Now let’s apply vector functions to a real-world scenario: understanding how gravity affects objects moving through the air. This situation\u2014where an object moves with an initial velocity but experiences only the force of gravity\u2014is called projectile motion<\/strong>. It describes everything from golf balls and baseballs to arrows and cannonballs. We’ll ignore air resistance in our analysis.<\/p>\n

        Setting Up the Coordinate System<\/h3>\n

        We place ourselves at the origin with the positive [latex]y[\/latex]-axis pointing up, the negative [latex]y[\/latex]-axis pointing down, and the positive [latex]x[\/latex]-axis pointing forward (away from the thrower). Since gravity acts downward, Newton’s second law tells us the force from gravity equals the mass times the acceleration due to gravity: [latex]F_{g}=mg[\/latex], where [latex]F_{g}[\/latex] is the gravitational force and [latex]g[\/latex] is the acceleration due to gravity at Earth’s surface.<\/p>\n

        The value of [latex]g[\/latex] is approximately [latex]32 ft\/sec^{2}[\/latex] in the English system and approximately [latex]9.8 m\/sec^{2}[\/latex] in the metric system. Since gravity acts downward, we write the gravitational force as [latex]F_{g}=-mg{\\bf{j}}[\/latex], as shown in Figure 1.<\/p>\n

        \"This
        Figure 1. An object is falling under the influence of gravity.<\/figcaption><\/figure>\n

        Deriving the Position Equation<\/h3>\n

        Newton’s second law states that [latex]F=m{\\bf{a}}[\/latex], where [latex]{\\bf{a}}[\/latex] is the acceleration vector. This force must equal the gravitational force at all times:<\/p>\n

        [latex]\\begin{array}{ccc}\\hfill {F} & =\\hfill & {F_{g}} \\hfill \\\\ \\hfill {m{\\bf{a}}}& =\\hfill & {-mg{\\bf{j}}} \\hfill \\\\ \\hfill {{\\bf{a}}}& =\\hfill & {-g{\\bf{j}}}\\hfill \\end{array}[\/latex]<\/p>\n

        Since the acceleration vector is the derivative of the velocity vector, we can write:<\/p>\n

        [latex]{{\\bf{v}}'(t)=-g{\\bf{j}}}[\/latex]<\/p>\n

        Taking the antiderivative of both sides:<\/p>\n

        [latex]\\begin{array}{ccc}\\hfill {{\\bf{v}}(t)} & =\\hfill & {\\int -g{\\bf{j}}dt} \\hfill \\\\ \\hfill & =\\hfill & {-gt{\\bf{j}}+{\\bf{C}}_{1}}\\hfill \\end{array}[\/latex]<\/p>\n

        To find the constant vector [latex]{\\bf{C}}{1}[\/latex], we use the initial velocity<\/strong> [latex]{\\bf{v}}{0}={\\bf{v}}(0)[\/latex]. This gives us [latex]{\\bf{C}}{1}={\\bf{v}}{0}[\/latex], so:<\/p>\n

        [latex]{\\bf{v}}(t)=-gt{\\bf{j}}+{\\bf{v}}_{0}[\/latex]<\/p>\n

        Since velocity is the derivative of position, we have:<\/p>\n

        [latex]{\\bf{s}}'(t)=-gt{\\bf{j}}+{\\bf{v}}_{0}[\/latex]<\/p>\n

        Taking the antiderivative again:<\/p>\n

        [latex]\\begin{array}{ccc}\\hfill {{\\bf{s}}(t)} & =\\hfill & {\\int -gt{\\bf{j}}+{\\bf{v}}{0}dt} \\hfill \\\\ \\hfill & =\\hfill & {-\\frac{1}{2}gt^{2}{\\bf{j}}+{\\bf{v}}{0}t+{\\bf{C}}_{2}}\\hfill \\end{array}[\/latex]<\/p>\n

        Using the initial position<\/strong> [latex]{\\bf{s}}{0}={\\bf{s}}(0)[\/latex], we find [latex]{\\bf{C}}{2}={\\bf{s}}_{0}[\/latex]. This gives the position at any time:<\/p>\n

        [latex]{\\bf{s}}(t)=-\\frac{1}{2}gt^{2}{\\bf{j}}+{\\bf{v}}{0}t+{\\bf{s}}{0}[\/latex]<\/p>\n

        Projectile Launched at an Angle<\/h3>\n

        Suppose an object is thrown upward from the origin at an angle [latex]\\theta[\/latex] to the horizontal with initial speed [latex]v_{0}[\/latex]. We can assume [latex]{\\bf{s}}_{0}={\\bf{0}}[\/latex] by placing the origin at the launch point (Figure 2).<\/p>\n

        \"This
        Figure 2. Projectile motion when the object is thrown upward at an angle [latex]\\theta[\/latex]. The horizontal motion is at constant velocity and the vertical motion is at constant acceleration.<\/figcaption><\/figure>\n

        We can express the initial velocity as [latex]{\\bf{v}}{0}=v{0}\\cos\\theta{\\bf{i}}+v_{0}\\sin\\theta{\\bf{j}}[\/latex]. The position equation becomes:<\/p>\n

        [latex]\\begin{array}{ccc}\\hfill {{\\bf{s}}(t)} & =\\hfill & {-\\frac{1}{2}gt^{2}{\\bf{j}}+v_{0}t\\cos\\theta{\\bf{i}}+v_{0}t\\sin\\theta{\\bf{j}}} \\hfill \\\\ \\hfill & =\\hfill & {v_{0}t\\cos\\theta{\\bf{i}}+(v_{0}t\\sin\\theta-\\frac{1}{2}gt^{2}){\\bf{j}}}\\hfill \\end{array}[\/latex]<\/p>\n

        \n

        components of projectile motion<\/h3>\n
          \n
        • The coefficient of [latex]{\\bf{i}}[\/latex] represents the horizontal distance<\/strong> from the origin at time [latex]t[\/latex]<\/li>\n
        • The maximum horizontal distance (at the same initial and final altitude) is called the range<\/strong> [latex]R[\/latex]<\/li>\n
        • The coefficient of [latex]{\\bf{j}}[\/latex] represents the altitude<\/strong> at time [latex]t[\/latex]<\/li>\n
        • The maximum altitude is called the height<\/strong> [latex]H[\/latex]<\/li>\n<\/ul>\n<\/section>\n
          During an Independence Day celebration, a cannonball is fired from a cannon on a cliff toward the water. The cannon is aimed at an angle of [latex]30^\\circ[\/latex] above horizontal and the initial speed of the cannonball is [latex]600\\ ft\/sec[\/latex]. The cliff is [latex]100\\ ft[\/latex] above the water (Figure 7).<\/p>\n
            \n
          1. Find the maximum height of the cannonball.<\/li>\n
          2. How long will it take for the cannonball to splash into the sea?<\/li>\n
          3. How far out to sea will the cannonball hit the water?
            \n
            \"This
            Figure 3. The flight of a cannonball (ignoring air resistance) is projectile motion.<\/figcaption><\/figure>\n<\/li>\n<\/ol>\n
            \n