Question

A 51.5-g golf ball is driven from the tee with an initial speed of 43.9 m/s and rises to a height of 33.9 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 5.33 m below its highest point?

Answer 1

the golf ball will go under projectile motion

m= 51.5 g

u = 43.9 m/s

Hmax= 33.9 m

applying energy conservation b.w initial and highest point

KEi + PEi = KEf + PEf

1/2 m u2 + 0 = KEf + mg Hmax

1/2 x 0.0515 x 43.9² = KEf +0.0515 x 9.81 x 33.9

49.62 = KEf + 17.13

KEf = 32.5 J

answer...a

b) let speed at height 5.33 m from the heighest point is V m/s

so from energy conservation

KEi + PEi = KEf + PEf

1/2 m u2 + 0 = KEf + mg H

H = 33.9 - 5.33 = 28.57 m

1/2 x 0.0515 x 43.9² = KEf +0.0515 x 9.81 x 28.57

49.62 = KEf + 14.43

KEf =35.18 J

1/2 m V² = 35.18

1/2 x 0.0515x V² = 35.18

V = 36.96 m/s