Most of us are aware of the general concept that the harder you hit something, the farther it usually travels. (let us suppose we are neglecting any air resistance and friction that happens) And from Newton, we have gotten the general formula $F=ma$ where force equals mass times acceleration.

Physics plays a big role in the sport of golf as well. And after thinking, I have been wondering which ball would travel farther. Suppose two golfers each hit a golf ball such that:

Golfer #1 hits the ball with a 10.5° driver of mass $m$ and hits the ball with a horizontal velocity $2v$

Golfer #2 hits the ball with a 10.5° driver of mass $2m$ and hits the ball with a horizontal velocity $v$

My question is which ball travels further horizontally? Or do they travel the same horizontal distance? Or is there not enough given information to tell?

I have tried figuring it out myself but haven't had any luck. I feel that acceleration is involved somewhere to link force to velocity (because acceleration is the derivative of velocity) but I'm not sure if this is the right approach.


Surprised to see, but no one has yet mentioned the logic of projectile here.

Since you've been given the angle of projection and the horizontal velocity of projection, the problem here becomes the classic case of projectile motion.

Note that the maximum horizontal distance $(R)$ that the golf ball can travel is


Where $v$ is the horizontal velocity by which it is projected, $g$ is the acceleration due to gravity and $sin2\theta$ is the sine of twice the angle of projection.

  • $\begingroup$ I was surprised too no one mentioned the case of a projectile, but well done, good answer @user38227 $\endgroup$ – Zlatan May 14 '17 at 4:38

In this case, it is the momentum that must be considered. Impulse is defined as the change in momentum of an object. The golf ball will always start with 0 momentum. If we assume that all of the clubs momentum is imparted on the ball (unlikely, but simplifies the math), then they both receive the same impulse. The impulse is also equal to force multiplied by the time interval; thus the balls experience the same force, and acceleration.

  • $\begingroup$ Ah I see. So therefore in this scenario the balls would travel the same distance? $\endgroup$ – WaveX Apr 20 '16 at 18:04
  • $\begingroup$ @WaveX Yes, as they will also have the same initial velocities. $\endgroup$ – Noah P Apr 20 '16 at 18:04
  • $\begingroup$ Okay. So my approach should have been towards momentum and not so much acceleration. Makes sense now. Thanks for your help $\endgroup$ – WaveX Apr 20 '16 at 18:05
  • $\begingroup$ @WaveX Yes, but in this case they are linked through the impulse. If you had worked backwards from acceleration, to force, to impulse, and then momentum, you would have reached the same answer. However as both the balls have an initial velocity of 0, then the acceleration would allow you to compare the final distance travelled. In this case the acceleration is identical,. $\endgroup$ – Noah P Apr 20 '16 at 18:08

If you neglect air resistance the mass of the golf ball does not influence the range with same initial velocity , only with air resistance it does.

But if you hit the ball the kinetic energy you transfer gets divided through a smaller mass, so the velocity of the lighter ball should be higher if hit both balls the same way:

$$e_{kin}=m\cdot v^2/2 \to v=\sqrt{2\cdot e_{kin}/{m}}$$

So in your scenario the lighter ball will travel farther.

Besides the initial velocity and angle it is mostly the Magnus-effect which is also used in airsoft-guns to increase the range, but this only works with air resistance.


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