Set up the equation of motion.

Okay, lekeufackshanif! Let's move on to Question 2, part a). Step 1: Set up the equation of motion. The forces acting on the ball are gravity downwards (mg) and resistive force upwards (mg (v^2)/(c^2)). Using Newton's Second Law, F_net = ma: m(dv)/(dt) = mg - mg(v^2)/(c^2) Divide by m: (dv)/(dt) = g(1 - (v^2)/(c^2)) Step 2: Separate variables and integrate. Rearrange the equation to separate v and t: (dv)/(1 - v^2)c^2 = g\,dt (c^2\,dv)/(c^2 - v^2) = g\,dt Integrate both sides. The left side is a standard integral of the form (1)/(a^2 - x^2) dx = (1)/(2a) | (a+x)/(a-x) |. Here, a=c and x=v. (c^2)/(c^2 - v^2)\,dv = g\,dt c^2 · (1)/(2c) | (c+v)/(c-v) | = gt + K (c)/(2) | (c+v)/(c-v) | = gt + K Step 3: Apply initial conditions to find the constant of integration K. The ball is dropped, so at t=0, v=0. (c)/(2) | (c+0)/(c-0) | = g(0) + K (c)/(2) (1) = K K = 0 Step 4: Solve for time t. Substitute K=0 back into the equation: (c)/(2) ( (c+v)/(c-v) ) = gt (We can remove the absolute value since v < c for the ball to be accelerating, so c+v and c-v are positive.) t = (c)/(2g) ( (c+v)/(c-v) ) Step 5: Calculate the time taken to attain a speed of (c)/(2). Substitute v = (c)/(2) into the expression for t: t = (c)/(2g) ( (c + c)/(2)c - (c)/(2) ) t = (c)/(2g) ( (3c)/(2)(c)/(2) ) t = (c)/(2g) (3) The time taken for the ball to attain a speed of (c)/(2) is (c)/(2g) (3). Last free one today — make it count tomorrow, or type /upgrade for unlimited.