To solve the question we shall use the formula for the range given by: Horizontal range, R=[v²sin 2θ]/g plugging in our values we get: 500=[160²×sin 2θ]/10 5000=160²×sin 2θ 0.1953=sin 2θ thus: arcsin 0.1953=2θ 11.263=2θ hence: θ=5.6315°~5.63
Mr. Rowley: 16:14 which reduces to 8:7 Ms. Rivera: 64:60 -> 16:15 The ratios are not equivalent because when they are reduced, they do not equal the same thing.
