A baseball is thrown straight up. The drag force is proportional to v^2.
In terms of g, what is the y-component of the ball's acceleration when its speed is half its terminal speed and it is moving up?
In terms of g, what is the y-component of the ball's acceleration when its speed is half its terminal speed and it is moving back down?
To solve this question, we need to consider the forces acting on the baseball and apply Newton's second law of motion.
1. When the ball is moving up and its speed is half the terminal speed:
The upward acceleration consists of two components: gravity and drag force. The drag force acting in the opposite direction of motion will reduce the net acceleration. The equation for drag force is given as F_drag = -kv^2, where k is a proportionality constant, and v is the velocity.
Let's denote the y-component of the acceleration as a_y. When the ball is moving up, the acceleration due to gravity is in the downward direction and is given by a_gravity = -g.
To find the acceleration due to the drag force, we need to determine the velocity of the ball. Let's assume the terminal velocity is v_terminal. When the speed is half the terminal speed, the velocity is v = (1/2)v_terminal.
The net acceleration is the sum of the acceleration due to gravity and the acceleration due to the drag force. Since the ball is moving up, the drag force will act in the opposite direction, so we have:
a_y = a_gravity + a_drag
= -g + (-k(1/2)v_terminal^2)
We can rewrite the terminal velocity in terms of g using the equation for terminal velocity of a falling object:
v_terminal = √(mg/k)
Substituting this into the equation for acceleration due to drag force, we get:
a_y = -g + (-k(1/2)(√(mg/k))^2)
= -g - (k/2)(mg/k)
= -g - (1/2)mg
Therefore, the y-component of the ball's acceleration when its speed is half its terminal speed, and it is moving up is given by -g - (1/2)mg.
2. When the ball is moving back down and its speed is half the terminal speed:
In this case, the acceleration due to gravity and the acceleration due to the drag force will have opposite directions. The equations remain the same, except that the sign of the gravitational acceleration becomes positive, as the ball is moving in the downward direction. Hence, the y-component of the ball's acceleration when it is moving down is:
a_y = a_gravity + a_drag
= g - k(1/2)v_terminal^2
= g - (1/2)mg
Therefore, the y-component of the ball's acceleration when its speed is half its terminal speed, and it is moving down is g - (1/2)mg.