1. **Problem 1: Signs 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, and she is moving toward Team USA net, so displacement is positive. (c) Acceleration vector sign: She is slowing down while moving toward Team USA net, 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 toward Team USA net, so displacement magnitude is increasing. 2. **Problem 2: Net displacement of Tyler McGregor** Tyler moves 5 m right (+x), 8 m forward (+y), then 2 m left (-x). Net displacement components: $$x = 5 - 2 = 3 \text{ m}$$ $$y = 8 \text{ m}$$ Magnitude of net displacement: $$d = \sqrt{x^2 + y^2} = \sqrt{3^2 + 8^2} = \sqrt{9 + 64} = \sqrt{73} \approx 8.54 \text{ m}$$ Direction (angle from x-axis): $$\theta = \tan^{-1}\left(\frac{8}{3}\right) \approx 69.44^\circ$$ 3. **Problem 3: Projectile motion of football kicked by Andres Borregales** Given: Initial velocity $v_0 = 25$ m/s Angle $\theta = 44^\circ$ Acceleration due to gravity $g = 9.81$ m/s$^2$ (a) Maximal height $H$: Vertical component of initial velocity: $$v_{0y} = v_0 \sin \theta = 25 \times \sin 44^\circ \approx 25 \times 0.6947 = 17.37 \text{ m/s}$$ At max height, vertical velocity is zero, so: $$H = \frac{v_{0y}^2}{2g} = \frac{(17.37)^2}{2 \times 9.81} = \frac{301.7}{19.62} \approx 15.38 \text{ m}$$ (b) Horizontal displacement (range) $R$: Horizontal component of initial velocity: $$v_{0x} = v_0 \cos \theta = 25 \times \cos 44^\circ \approx 25 \times 0.7193 = 17.98 \text{ m/s}$$ Time of flight $T$: $$T = \frac{2 v_{0y}}{g} = \frac{2 \times 17.37}{9.81} = \frac{34.74}{9.81} \approx 3.54 \text{ s}$$ Range: $$R = v_{0x} \times T = 17.98 \times 3.54 \approx 63.65 \text{ m}$$ 4. **Problem 4: Tennis ball bounce velocity and height** Given: Mass $m = 0.058$ kg (not needed for velocity/height) Drop height $h = 1.4$ m Velocity before impact $v_1 = 3$ m/s downward Coefficient of restitution $e = 0.76$ (a) Velocity after impact $v_2$: By definition: $$e = \frac{|v_2 - 0|}{|0 - (-v_1)|} = \frac{|v_2|}{|v_1|} \Rightarrow v_2 = e v_1 = 0.76 \times 3 = 2.28 \text{ m/s (upward)}$$ (b) Bounce height $h_b$: Using energy relation: $$e = \sqrt{\frac{h_b}{h}} \Rightarrow h_b = e^2 h = (0.76)^2 \times 1.4 = 0.5776 \times 1.4 = 0.8086 \text{ m}$$ --- **Final answers:** 1. (a) +, (b) +, (c) -, (d) decreasing, (e) increasing 2. Net displacement $\approx 8.54$ m at $69.44^\circ$ from right (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 $0.81$ m