Chapter 5: Problem 40
You throw a baseball straight upward. The drag force is proportional to \(v^2\). In terms of \(g\), what is the y-component of the ball's acceleration when the ball's speed is half its terminal speed and (a) it is moving up? (b) It is moving back down?
Short Answer
Expert verified
(a) \( \frac{5g}{4} \) when moving up; (b) \( \frac{3g}{4} \) when moving down.
Step by step solution
01
Understand the Forces Involved
When a baseball is thrown upwards, two main forces act on it: gravitational force, which is constant and equals to the weight of the ball \( F_g = mg \), and the drag force, which depends on the velocity squared, given by \( F_d = kv^2 \), where \( k \) is a drag coefficient specific to the situation.
02
Understand Terminal Velocity
Terminal velocity is the constant speed that a freely falling object eventually reaches when the resistance of the medium prevents further acceleration. At terminal velocity, the drag force equals the gravitational force: \( kv_t^2 = mg \). It follows that \( k = \frac{mg}{v_t^2} \).
03
Half Terminal Velocity Scenario
If the speed of the ball is half of its terminal speed, then \( v = \frac{v_t}{2} \). Substituting this into the drag force equation gives: \[ F_d = k\left(\frac{v_t}{2}\right)^2 = \frac{mg}{v_t^2}\left(\frac{v_t^2}{4}\right) = \frac{mg}{4} \].
04
Calculate Net Force When Moving Up
When the ball is moving up, the gravitational force and the drag force act in the same direction (downwards). Thus, the net force is: \[ F_{net} = mg + \frac{mg}{4} = \frac{5mg}{4} \].
05
Calculate Acceleration When Moving Up
Using Newton's second law, \( F = ma \), the y-component of the ball's acceleration when moving up is:\[ a = \frac{F_{net}}{m} = \frac{5mg/4}{m} = \frac{5g}{4} \].
06
Calculate Net Force When Moving Down
When the ball is moving down, the gravitational force acts downwards, while the drag force acts upwards. Therefore, the net force is:\[ F_{net} = mg - \frac{mg}{4} = \frac{3mg}{4} \].
07
Calculate Acceleration When Moving Down
Using Newton's second law, the y-component of the ball's acceleration when moving down is:\[ a = \frac{F_{net}}{m} = \frac{3mg/4}{m} = \frac{3g}{4} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Projectile Motion
Projectile motion occurs when an object is thrown into the air and moves under the influence of gravity. In this context, the motion is only in the vertical plane since the ball is thrown straight up. We must consider how the object moves up, reaches a peak height, and then moves back down. The constant force of gravity acts on the object throughout all of these stages, pulling it toward the earth. In this specific problem, analyzing the motion helps us understand the forces on the baseball at different points in its trajectory.
Drag Force
Drag force is a resistance force caused by the air in this scenario, acting opposite to the motion. It increases with the square of the velocity: the larger the speed, the greater the drag force. The formula for the drag force is given by:
- \( F_d = kv^2 \)
Terminal Velocity
Terminal velocity is the maximum velocity an object can reach as it falls through a fluid, like air, when forces become balanced. As an object falls, it accelerates due to gravity. Eventually, the drag force due to air resistance increases to equal gravitational force. This results in no net force, so the object stops accelerating and continues to fall at constant speed, i.e., terminal velocity. For the baseball:
- At terminal velocity, \( kv_t^2 = mg \).
Newton's Second Law
Newton's Second Law states that the net force acting on an object is equal to the mass of that object multiplied by its acceleration: \( F = ma \). This principle is essential to solve any motion and force-related problems. In the context of this projectile motion, it helps to calculate the y-component of acceleration:
- When moving up, both gravitational and drag forces act downwards.
- When moving down, gravity pulls down and drag acts upwards.