Question

A golf ball is launched with the initial conditions shown in the figure. Determine the radius of curvature of the trajectory and the time rate of change of the speed of the ball (a) just after launch and (b) at apex. Neglect aerodynamic drag 187 mi/hr 110 Answers: (a) ρ= ft, ft/sec2 归 ft, ft/sec2

Answer 1

From draw the diagram The acceleration due to gravity has two components. One along the normal direction and other along the tangential direction is agcos11" g cos11 g cos11 The initial velocity of the golf ball is, mi ( 5280 ft ( 1 hr hr1 mi 3600 s 274.27 ft/s Hence, the radius of curvature is, g cos11 274.27- 32.2 ft/s cos11 2379.86 ft The time rate of change of speed is nothing but, the tangential component of acceleration agsin 11 v-gsin 11 -- 32.2 ft/s )sin 11 -6.14 ft/sec From draw the diagram

The acceleration in the tangential direction zero. Ther efore the time rate of change of speed is, "zero" From the above figure, The normal component of acceleration is, an-g 187cos11) 187(5280) 3600 32.2 7.87x10 f/s The time rate of change of speed is nothing but, the tangential component of acceleration q, =-gsin 110 =-(32.2 ft/s*) sin 110 --6.14 ft/sec