1. **State the problems:** **Question 2:** Given the vector equation $v_x(30 - 10t) = 20v_y$ describing projectile motion components, a) Find initial horizontal and vertical components at $t=0$. b) Find time of flight. c) Find velocity on hitting the horizontal ground. d) Find speed after 1.5 seconds. **Question 3:** A model aircraft flies in horizontal circle of radius $r$ with tension $F$ and speed $v$, taking time $T$ for one circuit. a) Find new time to complete a circuit if string length is increased to $4r$ at same speed. b) Find new tension if length is $4r$, speed $v$. c) Find tension if length is $r$ but speed is increased to $4v$. d) Draw diagram showing velocity, tension, acceleration directions. --- 2. **Solve Question 2:** **a)** At $t=0$, input into $v_x(30 - 10t) = 20v_y$: $$v_x(30 - 0) = 20v_y \Rightarrow 30v_x = 20v_y \Rightarrow v_y = \frac{30}{20} v_x = 1.5 v_x$$ The initial horizontal component is $v_x$ (unknown constant), vertical component is $v_y = 1.5 v_x$. **b)** Time of flight is when projectile hits ground. For vertical motion under gravity $g$, if initial vertical velocity is $v_{y0}=1.5 v_x$, and assuming upward positive and standard projectile equations, time of flight $t_f = \frac{2 v_{y0}}{g} = \frac{2 \times 1.5 v_x}{9.8} = \frac{3 v_x}{9.8}$ seconds. **c)** Velocity hitting ground: Horizontal velocity remains $v_x$ (no horizontal acceleration). Vertical velocity at impact is $v_{yf} = -v_{y0}$ (downward), magnitude $1.5 v_x$. Magnitude of velocity: $$v = \sqrt{v_x^2 + (-1.5 v_x)^2} = v_x \sqrt{1 + 2.25} = v_x \sqrt{3.25} \approx 1.803 v_x$$ **d)** Speed after 1.5 seconds: Vertical velocity after time $t$: $$v_y = v_{y0} - g t = 1.5 v_x - 9.8 \times 1.5 = 1.5 v_x - 14.7$$ Horizontal velocity is still $v_x$. Speed: $$s = \sqrt{v_x^2 + (1.5 v_x - 14.7)^2}$$ --- 3. **Solve Question 3:** **a)** Original circumference $C = 2\pi r$, time $T = \frac{C}{v} = \frac{2\pi r}{v}$. New radius $4r$, circumference $C' = 2 \pi \times 4r = 8\pi r$. Time at same speed: $$T' = \frac{C'}{v} = \frac{8\pi r}{v} = 4 \times \frac{2\pi r}{v} = 4T$$ **b)** Tension providing centripetal force: $$F = m \frac{v^2}{r}$$ New radius $4r$, same $v$, new tension: $$F' = m \frac{v^2}{4r} = \frac{1}{4} F$$ **c)** Speed increased to $4v$, radius $r$: New tension: $$F'' = m \frac{(4v)^2}{r} = m \frac{16 v^2}{r} = 16F$$ **d)** Diagram description: - Circle center with radius $r$ to aircraft position. - Velocity vector tangential to circle direction. - Tension force vector points along string toward center. - Acceleration vector (centripetal) points toward center. --- **Final answers summary:** - Q2a: $v_y = 1.5 v_x$ at $t=0$. - Q2b: $t_f = \frac{3 v_x}{9.8}$. - Q2c: Impact speed $\approx 1.803 v_x$. - Q2d: Speed at $1.5s$ is $\sqrt{v_x^2 + (1.5 v_x - 14.7)^2}$. - Q3a: New time $4T$. - Q3b: New tension $\frac{1}{4} F$. - Q3c: New tension $16 F$. - Q3d: Diagram shows velocity tangent, tension and acceleration toward center.