At low speeds (especially in liquids rather than gases), the drag force is proportional to the speed rather than it's square, i.e., F⃗ = −C1rv⃗ , where C1 is a constant. At time t = 0, a small ball of mass m is projected into a liquid so that it initially has a horizontal velocity of u in the +x 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 C1, r, m, g, vx, vy, u and t (enter C_1 for C1, v_x for vx and v_y for vy). Enter e^(-z) for exp(-z) (the exponential function of argument -z).
(1)what is the acceleration in the y direction as a function of the component of the velocity in the y direction vy? express your answer in terms of vy, C1, r, g, m and u as needed:
ay=
(2) Using your result from part (a), find an expression for the horizontal component of the ball's velocity as a function of time t? express your answer in terms of C1, r, g, m, u and t as needed: (enter e^(-z) for exp(-z)).
vx(t)=
(3) Using your result from part (b), find an expression for the vertical component of the ball's velocity as a function of time t? express your answer in terms of C1, r, g, m, u and t as needed: (enter e^(-z) for exp(-z)).
vy(t)=
(4) How long does it take for the vertical speed to reach 99% of its maximum value? express your answer in terms of C−1, r, g, m and u as needed:
please anybody help......
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)
Did you get number 6?
(5*R)/2
did you got the vertical spring one?
To solve these questions, we will use the equation F⃗ = −C1rv⃗ , where F⃗ is the drag force, C1 is a constant, r is the velocity vector, and v⃗ is the velocity vector. We will also use the kinematic equations for motion in the y direction.
(1) The acceleration in the y direction, ay, can be obtained by equating the drag force with the gravitational force:
Drag force, F⃗ = −C1rv⃗
Gravitational force, mg = ma⃗
Since the object is moving horizontally, the only y-component of the acceleration is due to the drag force. Thus, we can equate the magnitude of the drag force to the mass times the y-component of acceleration:
C1rvy = mg
Rearranging the equation, we get:
ay = (mg) / (C1rvy)
So, the acceleration in the y direction as a function of the component of velocity in the y direction, vy, is:
ay = (mg) / (C1rvy)
(2) To find the expression for the horizontal component of the ball's velocity as a function of time, vx(t), we know that there is no drag force acting in the x direction. Therefore, the horizontal component of velocity remains constant, equal to the initial horizontal velocity, u:
vx(t) = u
(3) To find the expression for the vertical component of the ball's velocity as a function of time, vy(t), we can use the kinematic equation:
vfy = v0y + ayt
vfy represents the final velocity in the y direction, v0y is the initial velocity in the y direction (which is zero in this case), a is the acceleration in the y direction (from part 1), y is the vertical position (not given in the question), and t is the time.
Since v0y = 0, the equation becomes:
vy(t) = ayt
Substituting the value of ay from part 1, we get:
vy(t) = [(mg) / (C1rvy)] t
(4) To find the time it takes for the vertical speed to reach 99% of its maximum value, we need to find the time when vy(t) is equal to 0.99 times the maximum vertical speed, vmax.
0.99vmax = [(mg) / (C1rvy)] t
Solving for t:
t = (0.99vmax) / [(mg) / (C1rvy)]
Note: The maximum vertical speed is not provided in the question. If you have that information, you can substitute it into the above equation to find the time it takes to reach 99% of the maximum speed.