Ch 08: Dynamics II: Motion in a Plane

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 08: Dynamics II: Motion in a Plane

Problem 68a

Chapter 8, Problem 68a

In the absence of air resistance, a projectile that lands at the elevation from which it was launched achieves maximum range when launched at a 45° angle. Suppose a projectile of mass m is launched with speed into a headwind that exerts a constant, horizontal retarding force ${\vec{F}}_{wind} = - F_{wind} \hat{ı} .$ . Find an expression for the angle at which the range is maximum.

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Start by analyzing the forces acting on the projectile. The headwind exerts a constant horizontal retarding force $ F_{\text{wind}} = -F_{\text{wind}} \hat{i} $, which affects the horizontal motion of the projectile. The vertical motion is influenced only by gravity $ g $.

Write the equations of motion for the horizontal and vertical directions. For the horizontal motion, the acceleration is $ a_x = -\frac{F_{\text{wind}}}{m} $, and the velocity in the horizontal direction is $ v_x(t) = v_0 \cos(\theta) - \frac{F_{\text{wind}}}{m}t $. For the vertical motion, the acceleration is $ a_y = -g $, and the velocity in the vertical direction is $ v_y(t) = v_0 \sin(\theta) - g t $.

Determine the time of flight $ T $ by finding when the projectile returns to its initial elevation. Use the vertical displacement equation $ y(t) = v_0 \sin(\theta)t - \frac{1}{2}g t^2 $. Setting $ y(T) = 0 $, solve for $ T $: $ T = \frac{2v_0 \sin(\theta)}{g} $.

Find the horizontal range $ R $ by integrating the horizontal velocity over the time of flight. The horizontal displacement is $ x(T) = \int_0^T \left(v_0 \cos(\theta) - \frac{F_{\text{wind}}}{m}t\right) dt $. Solve this integral to get $ R = v_0 \cos(\theta)T - \frac{F_{\text{wind}}}{2m}T^2 $. Substitute $ T = \frac{2v_0 \sin(\theta)}{g} $ into this expression.

To maximize the range $ R $, take the derivative of $ R $ with respect to $ \theta $ and set it equal to zero. Solve $ \frac{dR}{d\theta} = 0 $ to find the angle $ \theta $ that maximizes the range. This will involve balancing the effects of the headwind and gravity on the projectile's motion.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Projectile Motion

Projectile motion refers to the motion of an object that is launched into the air and is subject to gravitational force. The trajectory of a projectile is typically parabolic, and its range is influenced by the launch angle, initial speed, and the effects of air resistance. In ideal conditions without air resistance, the optimal launch angle for maximum range is 45 degrees.

Effect of Air Resistance

Air resistance, or drag, is a force that opposes the motion of an object through the air. It affects the trajectory and range of projectiles by reducing their horizontal velocity and altering their optimal launch angle. In the presence of a headwind, the effective launch angle for achieving maximum range will differ from the ideal 45 degrees due to the additional retarding force acting on the projectile.

Optimal Launch Angle

The optimal launch angle is the angle at which a projectile should be launched to achieve the maximum horizontal distance or range. This angle can change based on external factors such as air resistance and wind. In the case of a projectile launched into a headwind, the optimal angle will need to be recalculated to account for the retarding force, which typically results in a launch angle less than 45 degrees.

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Textbook Question

In the absence of air resistance, a projectile that lands at the elevation from which it was launched achieves maximum range when launched at a 45° angle. Suppose a projectile of mass m is launched with speed into a headwind that exerts a constant, horizontal retarding force ${\vec{F}}_{wind} = - F_{wind} \hat{ı} .$ By what percentage is the maximum range of a 0.50 kg ball reduced if $F sub wind equals 0.60 cap N$ ?

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