A circus juggler performs an act with balls that he tosses with his right hand and catches with his left hand. Each ball is launched at an angle of \(75^{\circ}\) and reaches a maximum height of \(90 \mathrm{~cm}\) above the launching height. If it takes the juggler \(0.2 \mathrm{~s}\) to catch a ball with his left hand, pass it to his right hand and toss it back into the air, what is the maximum number of balls he can juggle?

We can use the kinematic equations for projectile motion to calculate the time of flight. We know the maximum height h and launch angle \(θ = 75°\). We'll start by finding the vertical component of velocity, \(v_y\): Using the equation: $$h = \frac{v^2_y}{2g}$$ Where h is the maximum height (0.9 m), \(v_y\) is the vertical velocity, and g is the acceleration due to gravity (approximately \(9.81 \mathrm{~m/s^2}\)). Solving for \(v_y\), we get: $$v_y = \sqrt{2gh} = \sqrt{(2)(9.81 \mathrm{~m/s^2})(0.9 \mathrm{~m})} ≈ 4.42 \mathrm{~m/s}$$ Now we'll find the total velocity at launch using the angle. We know that: $$v_y = v\sin(θ)$$ So, we can find v by: $$v = \frac{v_y}{\sin(θ)} = \frac{4.42 \mathrm{~m/s}}{\sin(75°)} ≈ 4.55 \mathrm{~m/s}$$ By knowing the total velocity, we can find the horizontal velocity component \(v_x\): $$v_x = v\cos(θ) ≈ (4.55 \mathrm{~m/s})(\cos(75°)) ≈ 1.18 \mathrm{~m/s}$$ Now we have both vertical and horizontal velocity components at launch. To find the time of flight, we'll use the equation: $$t = \frac{2v_y}{g}$$ Solving for t, we get: $$t = \frac{2(4.42 \mathrm{~m/s})}{9.81 \mathrm{~m/s^2}} ≈ 0.9 \mathrm{~s}$$ So, the time of flight for each ball is 0.9 s.

We know it takes 0.2 s for the juggler to catch, pass, and relaunch a ball. So, in the 0.9 s window where a ball is in flight, the juggler needs to have enough time to manage the other balls. Let n be the number of balls. The time taken to catch, pass, and relaunch the n-1 remaining balls must be less than the total time of flight of a single ball: $$(n-1)(0.2 \mathrm{~s}) ≤ 0.9 \mathrm{~s}$$ Solving for n: $$n-1 ≤ \frac{0.9 \mathrm{~s}}{0.2 \mathrm{~s}}$$ $$n-1 ≤ 4.5$$ $$n ≤ 5.5$$ Since the number of balls must be an integer, we can infer that the maximum number of balls the juggler can juggle is 5.