At low speeds (especially in liquids rather than gases), the drag force is proportional to the speed rather than it's square, i.e., F = -c_1*r*v , where c_1 is a constant. At time t = 0, a small ball of mass is projected into a liquid so that it initially has a horizontal velocity of in the direction as shown. The initial speed in the vertical direction ( y) is zero. The gravitational acceleration is g. Consider the cartesian coordinate system shown in the figure (+x to the right and +y downwards).
Express the answer of the following questions in terms of some or all of the variables c_1, r , m , g, v_x , v_y , and u (enter C_1 for , v_x for and v_y for ). Enter e^(-z) for exp(-z) (the exponential function of argument -z).
(a) What is component of the acceleration in the direction as a function of the component of the velocity in the direction ? express your answer in terms of v_x , c_1 , r , g , m and u as needed: What is the acceleration in the direction as a function of the component of the velocity in the direction ? express your answer in terms of v_y , c_1 , r , g, m and as needed: Using your result from part (a), find an expression for the horizontal component of the ball's velocity as a function of time? Using your result from part (b), find an expression for the vertical component of the ball's velocity as a function of time t. How long does it take for the vertical speed to reach 99 of its maximum value.What value does the horizontal component of the ball's velocity approach as t becomes infinitely large?What value does the vertical component of the ball's velocity approach as t becomes infinitely large?
a)(-C_1*r*v_x)/m
b)g-(C_1*r*v_y)/m
c)u*e^(-(C_1*r*t)/m)
d)((m*g)/(C_1*r))-((m*g)/(C_1*r))*e^(-(C_1*r*t)/m)
e)4.6*(m/(C_1*r))
f)0
g)(m*g)/(C_1*r)
I didn't agree with the answer
To find the component of acceleration in the x-direction, we need to consider the forces acting on the ball in that direction. The only force in the x-direction is the drag force, which is proportional to the velocity in the x-direction. Since the initial velocity in the x-direction is v_x, we can write the equation of motion in the x-direction as:
m * a_x = -c_1 * r * v_x
where m is the mass of the ball. Dividing both sides by m gives us the component of the acceleration in the x-direction:
a_x = -c_1 * r * v_x / m
Now, to find the acceleration in the y-direction, we need to consider the forces acting on the ball in that direction. The two forces acting on the ball in the y-direction are the drag force (proportional to the velocity in the y-direction) and the gravitational force (mg). We can write the equation of motion in the y-direction as:
m * a_y = -c_1 * r * v_y - m * g
where g is the acceleration due to gravity. Dividing both sides by m gives us the component of the acceleration in the y-direction:
a_y = -c_1 * r * v_y / m - g
Now that we have expressions for the acceleration in both the x-direction and y-direction, we can find the expressions for the horizontal and vertical components of the ball's velocity as a function of time.
To find the horizontal component of the ball's velocity, we integrate the expression for the acceleration in the x-direction with respect to time. Since the initial velocity in the x-direction is v_x, the initial condition for this integration would be:
v_x (t=0) = v_x
Integrating the expression for a_x gives us:
v_x(t) = v_x - (c_1 * r * v_x / m) * t
To find the vertical component of the ball's velocity, we integrate the expression for the acceleration in the y-direction with respect to time. Since the initial velocity in the y-direction is v_y, the initial condition for this integration would be:
v_y (t=0) = 0
Integrating the expression for a_y gives us:
v_y(t) = (c_1 * r * v_y / m - g) * t
To find the time it takes for the vertical speed to reach 99% of its maximum value, we can set the expression for v_y(t) to 0.99 times its maximum value (let's call it v_y_max) and solve for t:
0.99 * v_y_max = (c_1 * r * v_y_max / m - g) * t
Solving for t gives us:
t = 0.99 * v_y_max / ((c_1 * r * v_y_max / m - g))
As t becomes infinitely large, the horizontal component of the ball's velocity approaches v_x since the drag force is proportional to the velocity rather than its square.
As t becomes infinitely large, the vertical component of the ball's velocity approaches the terminal velocity, which is the maximum velocity it can reach in the presence of drag force. The terminal velocity can be found by setting the expression for a_y to 0 and solving for v_y:
0 = -c_1 * r * v_y_terminal / m - g
Solving for v_y_terminal gives us:
v_y_terminal = -m * g / (c_1 * r)