A baseball is thrown from the roof of a 22.0-m-tall building with an initial velocity of magnitude 12.0 m/s and directed at an angle of 53.1o above the horizontal. (a) What is the speed of the ball just before it strikes the ground? Use energy methods and ignore air resistance. (b) What is the answer for part (a) if the initial velocity is at an angle of 53.1o below the horizontal? (c) If the effects of air resistance are included, will part (a) or (b) give the higher speed?

Solution 5E The height from where the ball was thrown is 22 m. Initial velocity was, V 0 12 m/s. Angle is, = 53.1 .0 a) Initially the potential energy of the ball was, P.E initialmgh = m × 9.8 × 22 = 215.6 m J . To get the initial velocity which has two components, we need to use a little bit of trigonometric deductions. 0 V =x12 cos 53.1 = 7.205 V =y12 sin 53.1 = 9.6. The x component remains constant through the whole motion, so we need to consider the y component only. Because the x component will get cancelled as it is same everywhere. Initially the kinetic energy was, 2 2 2 K.E initial1/2 mV y = 1/2 × m × V y = 0.5 × m × 9.6 = 40 m. The final potential energy would be, P.E final mgh = 9.8 × m × h. Final kinetic energy is zero at the highest point where the ball reached. K.E final 0. We know from the conservation of mechanical energy that, P.E initialK.E initial P.E final+ K.E final 215.6 m + 72 m = 9.8 × m × h + 0 215.6 + 72...