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)
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?
To find the component of acceleration in the x-direction, we can use Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the only force acting on the ball in the x-direction is the drag force, given by F = -c_1 * r * v_x, where v_x is the component of velocity in the x-direction. Since F = ma, we can write the equation as:
-m * a_x = -c_1 * r * v_x
Simplifying, we get:
a_x = (c_1 * r * v_x) / m
So the component of acceleration in the x-direction is (c_1 * r * v_x) / m.
To find the component of acceleration in the y-direction, we need to consider two forces: gravity and drag. The force of gravity is given by F = m * g, where g is the acceleration due to gravity. The drag force in the y-direction is proportional to the velocity in the y-direction, which is initially zero. Therefore, the net force in the y-direction is simply the force of gravity.
-m * a_y = m * g
So the acceleration in the y-direction is -g.
Using the result from part (a), we can find the expression for the horizontal component of the ball's velocity as a function of time. Recall that acceleration is the rate of change of velocity, and since a_x is constant, we have:
a_x = dv_x/dt
Integrating both sides of the equation with respect to time, we get:
∫ a_x dt = ∫ dv_x
c_1 * r * t = v_x - u
where u is the initial velocity in the x-direction. Rearranging the equation, we find:
v_x = c_1 * r * t + u
So the horizontal component of the ball's velocity as a function of time is given by c_1 * r * t + u.
Using the result from part (b), we know that the acceleration in the y-direction is constant and equal to -g. Therefore, we can use the equations of motion to find the vertical component of the ball's velocity as a function of time. The equation of motion is:
v_y = u_y - g * t
Since the initial velocity in the y-direction is zero (u_y = 0), we have:
v_y = -g * t
To find the time it takes for the vertical speed to reach 99% of its maximum value, we set v_y equal to 0.99 times the maximum value:
- g * t = 0.99 * (initial vertical velocity)
Solving for t, we get:
t = (0.99 * initial vertical velocity) / g
As t becomes infinitely large, the horizontal component of the ball's velocity approaches the constant value of c_1 * r * t + u. So the answer is c_1 * r * t + u.
As t becomes infinitely large, the vertical component of the ball's velocity approaches the constant value of -g * t. So the answer is -g * t.