1. **Problem 1: Sign of vectors for Marie-Philip Poulin** Given: Positive direction is toward Team USA net. (a) Velocity vector sign: Since she is moving toward Team USA net, velocity is positive. (b) Displacement vector sign: Displacement is measured from Team Canada net toward Team USA net, so displacement is positive. (c) Acceleration vector sign: She is slowing down while moving forward, so acceleration is opposite to velocity, hence negative. (d) Velocity magnitude: Since she is slowing down, velocity magnitude is decreasing. (e) Displacement magnitude: She is still moving forward, so displacement magnitude is increasing. 2. **Problem 2: Tyler McGregor's net displacement** Tyler moves 5m right (+x), 8m forward (+y), then 2m left (-x). Net displacement components: $$x = 5 - 2 = 3\,m$$ $$y = 8\,m$$ Magnitude of displacement: $$d = \sqrt{x^2 + y^2} = \sqrt{3^2 + 8^2} = \sqrt{9 + 64} = \sqrt{73} \approx 8.54\,m$$ Direction (angle from x-axis): $$\theta = \tan^{-1}\left(\frac{8}{3}\right) \approx 69.44^\circ$$ 3. **Problem 3: Football kick by Andres Borregales** Given: Initial velocity $v = 25\,m/s$, angle $\theta = 44^\circ$, gravity $g = 9.81\,m/s^2$. (a) Maximal height $H$: Vertical component of velocity: $$v_y = v \sin \theta = 25 \times \sin 44^\circ \approx 25 \times 0.6947 = 17.37\,m/s$$ Use formula: $$H = \frac{v_y^2}{2g} = \frac{(17.37)^2}{2 \times 9.81} = \frac{301.7}{19.62} \approx 15.38\,m$$ (b) Horizontal displacement (range) $R$: Horizontal component of velocity: $$v_x = v \cos \theta = 25 \times \cos 44^\circ \approx 25 \times 0.7193 = 17.98\,m/s$$ Time to reach ground: Time to max height: $$t_{up} = \frac{v_y}{g} = \frac{17.37}{9.81} \approx 1.77\,s$$ Total flight time: $$t = 2 t_{up} = 3.54\,s$$ Range: $$R = v_x \times t = 17.98 \times 3.54 \approx 63.65\,m$$ 4. **Problem 4: Tennis ball bounce velocity and height** Given: Mass $m = 0.058\,kg$, drop height $h_d = 1.4\,m$, vertical velocity before impact $v_1 = 3\,m/s$ downward, coefficient of restitution $e = 0.76$. (a) Velocity after impact $v_2$: Coefficient of restitution formula: $$e = \frac{|v_2|}{|v_1|} \Rightarrow v_2 = e \times v_1 = 0.76 \times 3 = 2.28\,m/s$$ Direction is upward, so velocity after impact is $+2.28\,m/s$. (b) Bounce height $h_b$: Using energy relation: $$e = \sqrt{\frac{h_b}{h_d}} \Rightarrow h_b = e^2 \times h_d = (0.76)^2 \times 1.4 = 0.5776 \times 1.4 = 0.808\,m$$ **Final answers:** 1. (a) +, (b) +, (c) -, (d) decreasing, (e) increasing 2. Net displacement $\approx 8.54\,m$ at $69.44^\circ$ from x-axis 3. (a) Max height $\approx 15.38\,m$, (b) Range $\approx 63.65\,m$ 4. (a) Velocity after impact $2.28\,m/s$ upward, (b) Bounce height $0.808\,m$