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so lets say i have two spinning wheels above each other separated by a distance 'x' and I send a tennis ball between them- how do I find the distance the ball would go?

using Kinetic energy= Revolutions per minute

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  • $\begingroup$ Looks like you're thinking of using a pitching machine. Might want to research those. $\endgroup$
    – Carl Witthoft
    Oct 29, 2021 at 14:54

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The trajectory of a tennis ball ejected from a machine gets a bit complicated due to the air resistance drag which is of second order. The typical speeds are in the range of 30-50m/s. Also, some machines' wheels rotate at different speeds spinning the ball that makes a difference.

Let's say your turning wheels are shooting the balls out at a speed of 30m/s. Assuming a radius of 20 cm they have a circumference of 125.6cm=1.256m. They need to rotate at $$30/1.256=23.83rps =60*23.83=1432.4rpm$$

The machine has to throw the balls at an angle of between 4 to 10 degrees depending on the wind and spin. The range can be around 80feet.

trjactory of tenis ball

Here is an article on the subject, range of the ball

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Well, material limits tend to restrict edge-of-disk speeds to less than 1000 m/s. So what happens to a tennis ball if it is accelerated to Mach 2.9?

The stagnation temperature is about 540 F, so it will begin to melt, then explode. Plus the shock pressures would probably cause, um, spontaneous self disassembly.

So maybe take Mach 0.95 as a practical upper limit of launch speed that allows something resembling the thrown ball to return to earth.

The coefficient of drag can be taken to be 0.5 thanks to hundreds of experiments done on tennis balls by college kids.

The coefficient of lift will need to be iterated to find the best trajectory. But you probably want peak altitude of about 10% of the range and a launch angle of just a few degrees. Try 5 to start with.

Sadly, I couldn't find any record attempts for unrestricted tennis ball launchers. Something should be done about that. How about an X-Prize for the first 1500 m shot over level ground :)

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  • $\begingroup$ Someone's trying to be Randall's doppelganger! :-) $\endgroup$
    – Carl Witthoft
    Oct 29, 2021 at 14:55
  • $\begingroup$ Going form memory, I think our "potato cannons" sent tennis balls about 800 yards across "the mines" when I was a kid growing up in Oakdale Pa near Pittsburg. Powered by lighter fluid in a stack of metal tennis ball cans inside a PVC pipe - about the same distance as my Ben Pearson 35# pull bow. With proper spin control, you should get double the flight time from a wheel thrower. $\endgroup$
    – Phil Sweet
    Oct 29, 2021 at 19:32
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It depends.

The main two factors (I can think of) that affect it are:

  • the angle that the tennis ball is launched
  • if there is any sliding between the bass and the spinning wheels (this actually translates to the actual speed at launch and the rotational speed of the ball).

Additional factors are:

  • wind speed and direction
  • the $C_D$ coefficient of drag

If you really want to be pedantic, you need to also determine what latitude you are at to take account for the different value of g.

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  • $\begingroup$ Hi, Nmech wouldnt it also depend on the rpm of the wheels please? Any good sources to suggest please about the derivation. not sure where to start $\endgroup$
    – Henry
    Oct 28, 2021 at 15:36
  • $\begingroup$ @Henry - linear velocity of wheel at contact point = angular velocity x radius. if no slip, no significant compression of ball, Rwheel >> Rball, and symmetry between the wheels, then that wheel rim linear velocity becomes the launch velocity. the rest is a standard projectile problem. $\endgroup$
    – Pete W
    Oct 28, 2021 at 15:56
  • $\begingroup$ Hi, for linear velocity, would we times it by 2 please as there are two wheels? $\endgroup$
    – Henry
    Oct 28, 2021 at 15:59
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    $\begingroup$ no. if you throw a ball with two hands, does it go twice as fast as your hand at the time of release? Good luck $\endgroup$
    – Pete W
    Oct 28, 2021 at 16:01
  • $\begingroup$ thanks! that is good logic. when applying projectile motion please, how would one convert velocity to force, i was thinking F=ma, but then where would i find time from? $\endgroup$
    – Henry
    Oct 28, 2021 at 16:03

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