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).

(a) What is component of the acceleration in the x direction as a function of the component of the velocity in the x direction vx? express your answer in terms of vx, C1, r, g, m and u as needed:

ax=

(−1mC1⋅r)⋅vx
(b) 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=

(c) 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)=

u⋅e−tmC1⋅r
(d) 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)=

(e) 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:

t=

(f) What value does the horizontal component of the ball's velocity approach as t becomes infinitely large? express your answer in terms of C1, r, g, m and u as needed:

(g) What value does the vertical component of the ball's velocity approach as t becomes infinitely large? express your answer in terms of C1, r, g, m and u as needed

ax=(-C_1*r*v_x)/m

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)

To answer these questions, we need to apply Newton's second law to the motion of the ball in the x and y directions. We will use the formulas for acceleration that were given in the question, which are derived from the drag force being proportional to the speed rather than its square.

(a) The component of acceleration in the x-direction, ax, is given by the equation:

ax = (-C1 * r * vx) / m

(b) The acceleration in the y-direction, ay, is simply the gravitational acceleration, g, since there is no drag force acting in the vertical direction:

ay = g

(c) To find the horizontal component of the ball's velocity as a function of time, vx(t), we integrate the equation for acceleration with respect to time:

vx(t) = u * e^(-C1 * r * t / m)

(d) Similarly, to find the vertical component of the ball's velocity as a function of time, vy(t), we integrate the equation for acceleration with respect to time:

vy(t) = g * t

(e) To find the time it takes for the vertical speed to reach 99% of its maximum value, we equate vy(t) to 0.99 times the maximum value, which is g * t_max:

0.99*g*t_max = g*t

Simplifying, we find:

t = 0.99 * t_max

(f) As t becomes infinitely large, the horizontal component of the ball's velocity approaches the initial horizontal velocity, u.

(g) Similarly, as t becomes infinitely large, the vertical component of the ball's velocity approaches the terminal velocity, which is the maximum vertical velocity that can be achieved due to the drag force balancing the gravitational force. The terminal velocity can be found by setting the drag force equal to the gravitational force:

C1 * r * v_terminal = m * g

Solving for v_terminal, we get:

v_terminal = m * g / (C1 * r)

Therefore, the vertical component of the ball's velocity approaches v_terminal as t becomes infinitely large.