Here is a question I’ve been mulling over for a few days since I heard a cricket commentator raise it during the recent West Indies – South Africa Twenty20 match. How high do you need to hit a cricket ball in order for it to reach terminal velocity on its way down? – in other words, beyond what height does the height of the ball make no difference to the speed at which it hits the hands of the unlucky fielder underneath?
The commentator’s question raised a flurry of email responses, which were read out during the course of the match, some of which sounded somewhat bizarre to me.
Anyway, here is my take on it. A cricket ball is subject to air resistance which scales as its velocity squared. The relevant formula also has factors for the density of air, the ball’s radius squared, pi, and the ‘drag coefficient’ (divided by two). Putting some values in for a cricket ball, that gives me a force (in Newtons) of about 0.003 times the velocity squared (velocity in metres per second). At terminal velocity, this is equal to the weight of the ball (so drag = weight, and Newton’s first law says a balanced force gives no change in velocity, that is, terminal velocity has been reached.) The weight is about 1.6 Newtons, so equating the two means velocity is about 22 metres per second, or around 80 km an hour. Fancy taking that catch?
Now for the height. The acceleration due to gravity is about 10 metres per second, per second, which means that to get to 22 metres per second velocity we’d need a height of about 25 metres, and the ball would take about 2.2 seconds to fall. That’s not really all that high.
This is a very very rough estimate, but it gives you some idea. Of course, knowing this makes no difference whatsover to my chances of actually holding on to the catch.
Geoff says:
Interesting problem! At the NZ vs Pakistan test here in Dunedin last week we frequently saw the fast bowlers deliver at 139 kph (38.6 m/s). This is nearly twice your result for terminal velocity! I suspect you may have used the ball’s diameter in your calculations instead of its radius.
So I thought I’d have a crack, too. =) Hope what follows isn’t too pedantic! I just happen to be working a lot with cricket ball motion at the moment, which is how I came across your post.
According to Prandtl (1917) the resistive force on a spherical object can be approximated quadratically as:
F = 0.625 * p * r^2 * v^2
Where p (rho, excusing ascii limitations) is air density (approx 1.2 kg/m^3 at 20 degrees celcius), r is ball radius and v is velocity. You used a slightly different equation, but the results should be pretty much the same.
Equating the resistive force to a 156g cricket ball (with diameter 7.19cm) under gravity I get a terminal velocity of just under 40 m/s. If that’s the case, then those fast bowlers are most likely releasing the ball as fast as the local air density allows!
Apparently the drag co-efficient for a cricket ball varies quite a lot according to a whole bunch of scary fluid dynamics associated with turbulent flow. I don’t know how appropriate the number 0.625 is in this case.
So, err, this doesn’t make the prospect of catching a cricket ball travelling at terminal velocity any less frightening! I’ll stick with tennis balls and “one hand, one bounce” =)
As for the height calculation, that’s an exercise for anyone crazy enough to enjoy working with integrals. Bear in mind the net acceleration of the ball is a function of both gravity AND drag (which itself is a function of velocity squared).
And don’t forget that the horizontal component of motion will contribute, hence the height required to reach terminal velocity (which should be the maximum possible height minus the height at which the ball left the bat) decreases if you hit it further! I think… Now I’m confused, cos surely drag would damp the horizontal velocity such that the ball would reach the ground again before hitting terminal velocity… Unless it was hit more or less vertically.
Hehehe. Anyway, that’s my ten cents worth. Have fun!
Geoff
Marcus Wilson says:
“Whole bunch of scary fluid dynamics”. Yes, quite. I avoid it whenever possible.