physics
posted by Anonymous .
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:

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)
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