A golf ball is struck by a golf club at 55 m/s at an angle of 28.0o above the horizontal. The mass of the ball is 45 grams and is in contact with the golf club for 1.5 m/s. Calculate the average force exerted on the ball by the club.

Answer 1

a=(V-Vo)/t
a = (55-0)/1.5*10^-3=36670 m/s^2

F = m*a = 0.045 * 36,670 = 1650 N.

Answer 2

No. I believe the answer should be 1460 N Henry. The initial velocity of the club is striking the ball at a 28 degree angle and not straight on so the x component of velocity from the club would be 48.5621 m/s, which is from cos(28)*55m/s.

Answer 3

The force of the golf club acts in parallel with the motion of the ball
which is 28o. So my answer is 1650 N. @
28o. Since they did not ask for the
direction of the force, I did not include the angle. Your answer is not
the full force acting on the ball; it is
the hor. component of the force.

If the motion of the ball was hor., I
would use your procedure.