Projectile Motion

(a) Calculate how long it takes for the package to reach the highway.

• Given: \Delta dy = -350 \text{ m}$,$v{ix} = 52 \text{ m/s}$,$\theta = 0^\circ$</p></li><li><p><strong>Required</strong>:$\Delta t$</p></li><li><p><strong>Analysis</strong>: Since$\theta = 0^\circ$,$v{iy} = vi \sin 0^\circ = 0$. Use$\Delta dy = v{iy}\Delta t - \frac{1}{2} g(\Delta t)^2 \implies \Delta d_y = -\frac{1}{2} g(\Delta t)^2$</p></li><li><p><strong>Solution</strong>:$\Delta t = \sqrt{\frac{-2\Delta d_y}{g}} = \sqrt{\frac{-2(-350\text{ m})}{9.8\text{ m/s}^2}} = 8.45\text{ s} \approx 8.5\text{ s}$(rounded to two significant digits).</p></li></ul></li><li><p><strong>(b) Determine the range of the package.</strong></p><ul><li><p><strong>Given</strong>:$\Delta dy = -350 \text{ m}$,$vi = 52 \text{ m/s}$,$\Delta t = 8.45 \text{ s}$(from part a)</p></li><li><p><strong>Required</strong>:$\Delta d_x$</p></li><li><p><strong>Analysis</strong>: Use$\Delta dx = (vi \cos \theta)\Delta t$(since$\theta = 0^\circ$,$\cos \theta = 1$, so$\Delta dx = vi\Delta t$)</p></li><li><p><strong>Solution</strong>:$\Delta d_x = (52\text{ m/s})(8.45\text{ s}) = 439.4\text{ m} \approx 4.4 \times 10^2 \text{ m}$(rounded)</p></li></ul></li></ul><h4 id="a8ee08a3-1c86-470c-a12f-5b237503c683" data-toc-id="a8ee08a3-1c86-470c-a12f-5b237503c683" collapsed="false" seolevelmigrated="true">Sample Problem 2: Solving Projectile Motion Problems with an Initial Vertical Velocity</h4><ul><li><p><strong>Scenario</strong>: A golfer hits a ball with$v_i = 25 \text{ m/s}$at$\theta = 30.0^\circ$above the horizontal. The ball lands$14 \text{ m}$below its initial height.</p></li><li><p><strong>(a) Calculate the maximum height of the ball.</strong></p><ul><li><p><strong>Given</strong>:$v_i = 25 \text{ m/s}$,$\theta = 30.0^\circ$</p></li><li><p><strong>Required</strong>:$\Delta d_{y,max}$</p></li><li><p><strong>Analysis</strong>: At max height,$v{fy} = 0$. First find$\Delta t$to reach max height using$v{fy} = vi \sin \theta - g\Delta t$. Then use$\Delta dy = (v_i \sin \theta)\Delta t - \frac{1}{2} g(\Delta t)^2$.</p></li><li><p><strong>Solution</strong>:</p><ul><li><p>$0 = vi \sin \theta - g\Delta t \implies \Delta t = \frac{vi \sin \theta}{g} = \frac{(25\text{ m/s})\sin 30.0^\circ}{9.8\text{ m/s}^2} = 1.28\text{ s}$</p></li><li><p>$\Delta d_{y,max} = (25\text{ m/s})(\sin 30.0^\circ)(1.28\text{ s}) - \frac{1}{2}(9.8\text{ m/s}^2)(1.28\text{ s})^2 = 8.0\text{ m}$</p></li></ul></li></ul></li><li><p><strong>(b) Determine the ball’s velocity on landing.</strong></p><ul><li><p><strong>Given</strong>:$vi = 25 \text{ m/s}$,$\Delta dy = -14 \text{ m}$(relative to launch height),$\theta = 30.0^\circ$</p></li><li><p><strong>Required</strong>:$v_f$</p></li><li><p><strong>Analysis</strong>: Calculate final vertical velocity ($v{fy}$) using$v{fy}^2 = (vi \sin \theta)^2 - 2g\Delta dy$. Calculate constant horizontal velocity ($vx = vi \cos \theta$). Combine using$vf = \sqrt{vx^2 + v_{fy}^2}$and direction using inverse tangent.</p></li><li><p><strong>Solution</strong>:</p><ul><li><p>$v_{fy}^2 = (25\text{ m/s} \cdot \sin 30.0^\circ)^2 - 2(9.8\text{ m/s}^2)(-14\text{ m}) = (12.5\text{ m/s})^2 + 274.4\text{ m}^2\text{/s}^2 = 156.25 + 274.4 = 430.65$</p></li><li><p>$v_{fy} = -\sqrt{430.65} = -20.8\text{ m/s}$(negative because moving downwards)</p></li><li><p>$v_x = 25\text{ m/s} \cdot \cos 30.0^\circ = 21.7\text{ m/s}$</p></li><li><p>$v_f = \sqrt{(21.7\text{ m/s})^2 + (-20.8\text{ m/s})^2} = \sqrt{470.89 + 432.64} = \sqrt{903.53} = 30.1\text{ m/s}$</p></li><li><p>$\theta{\text{landing}} = \tan^{-1}\left(\frac{|v{fy}|}{|v_x|}\right) = \tan^{-1}\left(\frac{20.8}{21.7}\right) = 44^\circ$</p></li><li><p><strong>Statement</strong>: The velocity of the ball when it lands is$30.1 \text{ m/s} [44^\circ \text{ below the horizontal}]$.</p></li></ul></li></ul></li></ul><h4 id="bdcbf3e5-12c2-4730-a1a0-3d5326a1e870" data-toc-id="bdcbf3e5-12c2-4730-a1a0-3d5326a1e870" collapsed="false" seolevelmigrated="true">The Range Equation (Special Case: Landing at Same Height)</h4><ul><li><p>This equation applies specifically when a projectile lands at the <strong>same height</strong> from which it was launched ($\Delta d_y = 0$).</p></li><li><p><strong>Derivation of Time of Flight ($\Delta t$)</strong>:</p><ul><li><p>Start with$\Delta dy = (vi \sin \theta)\Delta t - \frac{1}{2} g(\Delta t)^2$</p></li><li><p>Set$\Delta dy = 0$:$0 = (vi \sin \theta)\Delta t - \frac{1}{2} g(\Delta t)^2$</p></li><li><p>Factor out$\Delta t$:$0 = \Delta t\left(v_i \sin \theta - \frac{1}{2} g\Delta t\right)$</p></li><li><p>This gives two solutions:$\Delta t = 0$(initial launch) or$v_i \sin \theta - \frac{1}{2} g\Delta t = 0$(landing).</p></li><li><p>Solving the second equation for$\Delta t$gives the time of flight for$\Delta dy=0$:$\Delta t = \frac{2 vi \sin \theta}{g}$</p></li></ul></li><li><p><strong>Derivation of Range ($\Delta d_x$)</strong>:</p><ul><li><p>Start with$\Delta dx = v{ix}\Delta t = (v_i \cos \theta)\Delta t$</p></li><li><p>Substitute the derived time of flight:$\Delta dx = (vi \cos \theta)\left(\frac{2 v_i \sin \theta}{g}\right)$</p></li><li><p>Rearrange:$\Delta dx = \frac{vi^2}{g} (2 \sin \theta \cos \theta)$</p></li><li><p>Using the trigonometric identity$2 \sin \theta \cos \theta = \sin(2\theta)$, the <strong>Range Equation</strong> is:<br>$\Delta dx = \frac{vi^2}{g} \sin(2\theta)$</p></li></ul></li><li><p><strong>Condition for application</strong>: This equation is valid <strong>only when$\Delta d_y = 0$</strong> (projectile lands at the same height it was launched from).</p></li><li><p><strong>Maximum Range</strong>: The largest value for the range occurs when$\sin(2\theta) = 1$. This happens when$2\theta = 90^\circ$, meaning$\theta = 45^\circ$. Therefore, the maximum range for a given initial speed is achieved at a launch angle of$45^\circ$$.

Air Resistance

• All discussions and examples of projectile motion assume that air resistance is negligible.

• This assumption is reasonably accurate for relatively dense objects moving at low speeds (e.g., a shot put).

• However, in many real-world situations, air resistance cannot be ignored, which makes the analysis of projectile motion significantly more complex and is beyond the scope of this text.