Question

Kramer decides to hit golf balls from the edge of a cliff into the ocean. The cliff is 62.9 m high. He hits the first ball (mass 45.9 g) giving it an initial velocity of 37.5 m/s at an angle of 22.1 above the horizontal. Determine:

a) the work done on the ball by gravity on the way up to its maximum height

b) the work done on the ball by gravity on the way down to the ocean

Answers: a) -4.56J b) 32.9J

Answer 1

(a) Work done on the ball by gravity on the way up to its maximum height which is given as :

maximum height, h = (v_o sin )² / 2g                                                  { eq.1 }

where, vo = initial velocity of ball = 37.5 m/s

= angle above horizontal = 22.1 degree

g = acceleration due to gravity = 9.8 m/s²

inserting all these values in above eq.

h = [(37.5 m/s) sin 22.1⁰]² / 2 (9.8 m/s²)

h = (14.1 m/s)² / (19.6 m/s²)

h = (198.8 m²/s²) / (19.6 m/s²)

h = 10.1 m

we know that, W = P.E

W_grav = -mg h                                                                       { eq.2 }

where, m = mass of the ball = 45.9 g = 0.0459 kg

inserting the values in eq.2,

W_grav = - (0.0459 kg) (9.8 m/s²) (10.1 - 0 )m

W_grav = -4.54 J

(b) Work done on the ball by gravity on the way down to the ocean which is given as :

total height, H = (62.9 m) + (10.1 m) = 73 m

using eq.2,     W_grav = -mg h    

inserting the values in above eq.

W_grav = - (0.0459 kg) (9.8 m/s²) (0 - 73)m

W_grav = 32.8 J