Question

A 48.8-g golf ball is driven from the tee with an initial speed of 46.7 m/s and rises to a height of 26.9 m. (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b) What is its speed when it is 6.43 m below its highest point?

Answer 1

mass of ball (m) = 48.8 g = 0.0488 kg

initial speed (u) = 46.7 m/s

so Total energy (TE) = ½mu² = 0.5 x 0.0488 x 46.7² = 53.213716 J

at the highest position gravitational potential energy (PE) = mgH = 0.0488 x 9.8 x 26.9 = 12.864656 J

(a) so kinetic energy at the highest position = TE - PE = 53.213716 - 12.864656 = 40.34906 J

(b) potential energy at 6.43 m below its highest point = mg(H-h) = 0.0488 x 9.8 x (26.9 - 6.43) = 9.7895728 J

so kinetic energy at that position (KE) = TE - PE = 53.213716 - 9.7895728 = 43.4241432 J

so ½mv² = 43.4241432

0.5 x 0.0488 x v² = 43.4241432

v²  = 1779.678

v = 42.186 m/s