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15c to find the time it takes the projectile to strike the ground … 58 1 z ′ ( t ) = − gt 2 + vD′ t sin α + ω vDt 2 cos α cos λ = 0 2 2vD′ sin α 2v′ sin α or t= ≈ D g − 2ω vD′ cos α cos λ g We have ignored the second term in the denominator—since vD′ would have to be impossibly large for the value of that term to approach the magnitude g g − 2ω vD′ cos α cos λ ≈ g − ω vD′ For example, for λ = 41o and α = 45o or vD′ ≈ g ω ≈ 144 km ! 22 Choose a coordinate system with the origin at the center of the wheel, the x′ and y′ axes pointing toward fixed points on the rim of the wheel, and the z ′ axis pointing toward the center of curvature of the track.

5 (a) ˆ + ˆjy F = ix ˆ + ˆjdy dr = idx on the path x = y : (1,1) ∫( 0,0 ) 1 1 1 1 0 0 0 0 F ⋅ dr = ∫ Fx dx + ∫ Fy dy = ∫ xdx + ∫ ydy = 1 ˆ dr = idx dr = ˆjdy on the path along the x-axis: and on the line x = 1 : (1,1) ∫( 0,0 ) 1 1 0 0 F ⋅ dr = ∫ Fx dx + ∫ Fy dy = 1 F is conservative. (b) ˆ − ˆjx F = iy on the path x = y : (1,1) ∫( 0,0 ) 1 1 0 0 1 1 0 0 F ⋅ dr = ∫ Fx dx + ∫ Fy dy = ∫ ydx − ∫ xdy (1,1) and, with x = y ∫( on the x-axis: ∫( and, with y = 0 on the x-axis ∫( on the line x = 1 : ∫( 0,0 ) (1,0 ) 0,0 ) (1,0 ) 0,0 ) (1,1) 1,0 ) 1 1 0 0 F ⋅ dr = ∫ xdx − ∫ ydy = 0 1 1 0 0 F ⋅ dr = ∫ Fx dx = ∫ ydx F ⋅ dr = 0 1 1 0 0 F ⋅ dr = ∫ Fy dy = ∫ xdy 35 (1,1) ∫( and, with x = 1 on this path 1,0 ) (1,1) ∫( 0,0 ) 1 F ⋅ dr = ∫ dy = 1 0 F ⋅ dr = 0 + 1 = 1 F is not conservative.

Is the distance behind home plate where the ball would be hypothetically launched at some angle α to achieve the total range R. x1 (=328 ft) is the distance the ball actually would travel from home plate if not caught by Mickey. (Note, because of the symmetry, v0 is the speed of the ball when it strikes the ground … also at the same angle α at which was launched. 9 zmax . 40 x1 Now solve for x0 using a relation identical to (4) … Thus, tan α = h0 = x0 ( x tan α ) tan α − 0 2 4 zmax Again we obtain a quadratic expression for u = x0 tan α which we solve as before.

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