Question

A golf ball is launched with the initial conditions shown in the figure. Determine the two times t1 2) when the radius of curvature of the trajectory has a value of 1665 ft. 164 mifhr Answers: t1 = 111.02294 t2 120.7655

Answer 1

Answer will be t₁ = 3.53 sec and t₂ = 1.333 sec

Below is my solution

- Radius of curvature of the trajectory: Nout. given o R= 1665ft se u Cose = 164 cos J9° = 155.065 milho => x= 355.065x1.466 ft/s » s = 227.43 ft/s. Ý = usino-yte = 164 sin 19894 = BS 53.993 milho - gt - 8 - 93-933 X + 466 4/5 - 8 → = 78-31 ft/s.- 32:27 Som Put į = -g - 322+1/s? A = 227.43 ft's, j = saya 78.31 ft/s - 32.27 üro , = 32.2 ft/s in ean 1665 = [122743) 6831 –32.2452738 => (227.43 X 322) (1219 3304-59) = (227-43)*+ (783) – 32,2x)? ) 1251513 = 6132.45 + 3036844 5043-1644 solving above quadratic egn, we got - t = 3.53 sec , 1.333 sec Answer to 5 930 t, - 3.530 sec 4,- 1.333 sec