Friction Golf
A golf ball with an initial angle of 34 ° lands exactly 226 m down the range on a level course.?
A. Denial of air friction, initial velocity, which would achieve this result? (m / s) B. Using the speed determined at point A, determine the maximum height reached the ball (m)
Let U projection speed, a is the angle to the horizontal, t is the flight time of the projection to landing x horizontal distance, the maximum height h, g is the acceleration of gravity. A. The horizontal and vertical resolution: 0 = no C (a) - gt ^ 2 / 2 ... (1) cos x = ut (a) ... (2) (1): (u T sin (a) - GT / 2) = 0 t = 0 (initially) landing: u sin (a) - GT / 2 = 0 t = 2u sin (a) / G ... (3) (2): x = t / u [cos (a)]. .. (4) Equating t (3) and (4): 2U sin (a) / g = x / cos u [(a)] 2u ^ 2 sin (a) cos (a) = gx u ^ 2 sin (2a) = gx u ^ 2 = GX / sin (2a) ... (5) u = sqrt [GX / sin (2a)] = sqrt [9.81 * 226 / sin (68)] = 48.9 m / s to 3SF. B. 0 ^ 2 = u ^ 2 - 2GH h = u ^ 2 / (2g) Substituting for u ^ 2 (5): h = x / [2 sin (2a)] = 226 / [2 sin (68)] = 121.9 m . to 3SF.
