youre 6.0 m from one wall of a house. you want to toss a ball to your friend who is 6.0 m from the opposite…

Explanation:

Step1: Analyze the horizontal and vertical - motion of the projectile

The horizontal distance between you and your friend is (d = 12.0\ m) (since you are 6.0 m from one wall and your friend is 6.0 m from the opposite wall), and the vertical height of the roof is (y=1.0\ m). The equations for horizontal and vertical motion of a projectile are (x = v_0\cos\theta t) and (y=v_0\sin\theta t-\frac{1}{2}gt^2). From (x = v_0\cos\theta t), we get (t=\frac{x}{v_0\cos\theta}). Substitute (t) into the vertical - motion equation: (y = v_0\sin\theta\frac{x}{v_0\cos\theta}-\frac{1}{2}g(\frac{x}{v_0\cos\theta})^2), which simplifies to (y=x\tan\theta-\frac{gx^{2}}{2v_0^{2}\cos^{2}\theta}).

Step2: For part A (find the minimum speed)

At the minimum speed, the ball just clears the roof. The horizontal distance (x = 12.0\ m) and (y = 1.0\ m). When the ball just clears the roof, we can use the fact that for a projectile, the optimal angle for maximum range for a given initial height and final height is considered. In the case of clearing an obstacle, we can also use the energy - conservation and kinematic approach. The minimum speed occurs when the ball is thrown at a (45^{\circ}) angle for the case of clearing an obstacle with equal starting and ending horizontal distances from the obstacle. The horizontal motion gives (x = v_0\cos\theta t), and the vertical motion gives (y=v_0\sin\theta t-\frac{1}{2}gt^2). Substituting (\theta = 45^{\circ}) ((\cos\theta=\sin\theta=\frac{\sqrt{2}}{2})), (x = 12.0\ m) and (y = 1.0\ m) into (x = v_0\cos\theta t) gives (t=\frac{x}{v_0\cos\theta}=\frac{12}{v_0\frac{\sqrt{2}}{2}}), and into (y=v_0\sin\theta t-\frac{1}{2}gt^2). [ \begin{align*} 1&=v_0\frac{\sqrt{2}}{2}\times\frac{12}{v_0\frac{\sqrt{2}}{2}}-\frac{1}{2}g(\frac{12}{v_0\frac{\sqrt{2}}{2}})^2\ 1&=12-\frac{1}{2}g\frac{144}{v_0^{2}\times\frac{1}{2}}\ \frac{1}{2}g\frac{288}{v_0^{2}}&=11\ v_0^{2}&=\frac{288g}{22}\ v_0&=\sqrt{\frac{288\times9.8}{22}}\approx11.4\ m/s \end{align*} ]

Step3: For part B (find the angle)

We know (x = 12.0\ m), (y = 1.0\ m), and from (y=x\tan\theta-\frac{gx^{2}}{2v_0^{2}\cos^{2}\theta}), using the identity (\frac{1}{\cos^{2}\theta}=1 + \tan^{2}\theta). Let (u = \tan\theta), then (y=xu-\frac{gx^{2}}{2v_0^{2}}(1 + u^{2})). Substituting (x = 12\ m), (y = 1\ m) and (v_0\approx11.4\ m/s) into (y=xu-\frac{gx^{2}}{2v_0^{2}}(1 + u^{2})): [ \begin{align*} 1&=12u-\frac{9.8\times144}{2\times(11.4)^{2}}(1 + u^{2})\ 1&=12u-\frac{1411.2}{2\times129.96}(1 + u^{2})\ 1&=12u - 5.4(1 + u^{2})\ 5.4u^{2}-12u + 6.4&=0 \end{align*} ] Using the quadratic formula (u=\frac{12\pm\sqrt{144 - 4\times5.4\times6.4}}{2\times5.4}=\frac{12\pm\sqrt{144 - 138.24}}{10.8}=\frac{12\pm\sqrt{5.76}}{10.8}=\frac{12\pm2.4}{10.8}). We get two solutions for (u=\tan\theta): (u_1=\frac{12 + 2.4}{10.8}\approx1.33) and (u_2=\frac{12 - 2.4}{10.8}\approx0.89). The smaller angle is (\theta=\arctan(0.89)\approx41.7^{\circ})

Answer:

Part A: (11.4\ m/s) Part B: (41.7^{\circ})