The classic physics problem. In a vacuum if all is totally horizontal (ground and throw trajectory) and gravity is perfectly perpendicular both should hit the ground at the same time.
BUT how does a atmosphere change the results? Drag is proportional to velocity^2, it wold seem at quick glance that the downward component’s would not be effected by the horizontal force of drag from the throw (initial horizontal velocity), and drag opposing the acceleration downward would be the same on both balls. However because it is falling in a parabola is there additional aerodynamic effects which might lead to lift (higher pressure below, lower above due to downward acceleration due to gravity which is higher then that of the dropped ball)?
Or in short which ball hits the ground first in a atmosphere and if there is a difference why?
If drag were proportional to velocity, the they would still hit at the same time. But because drag is proportional to v^2, the thrown ball will be slowed slightly more, and hence the dropped ball will hit slightly quicker.
This is assuming no asymmetries due to things like shape of the objects or spin. If those are preset, and speeds are relatively low, then those asymmetries will probably be more significant than the quadratic drag, and either result is possible depending on the details.
I can tell you for certain that a bullet fired from a gun and dropped will strike the ground at the same time. Mythbusters covered this in one of their episodes.
Back to the balls, though. Assuming they’re perfectly round and smooth, any aerodynamic forces (from the lateral movement) should be evenly balanced on top and bottom. If there were a ridge or similar imperfection (like a baseball, for example) that’d create a slight amount of lift (not enough to make it rise, just fall slower).
Detailed math: Assume that both balls have a vertical component of velocity of v[sub]y[/sub], and the thrown ball additionally has a horizontal component of v[sub]x[/sub]. The dropped ball has a vertical component of drag equal to cv[sub]y[/sub][sup]2[/sup], where c is a constant which will be the same for both balls. The thrown ball will have a total speed of sqrt(v[sub]x[/sub][sup]2[/sup] + v[sub]y[/sub][sup]2[/sup]) , for a total drag of c(v[sub]x[/sub][sup]2[/sup] + v[sub]y[/sub][sup]2[/sup]) , in the direction opposing the motion. The vertical component of the drag will be equal to the total drag times v[sub]y[/sub]/sqrt(v[sub]x[/sub][sup]2[/sup] + v[sub]y[/sub][sup]2[/sup]) , which reduces to c*v[sub]y[/sub]*v[sub]total[/sub]. Since v[sub]total[/sub] > v[sub]y[/sub], this means that, for the same vertical speed, the ball with horizontal motion will have greater drag than the ball without horizontal motion.
Careful here. Drag is proportional to the magnitude of the velocity vector squared. This means that the downward motion is affected by the horizontal motion, since the magnitude depends both on the horizontal and vertical components.
To see this, suppose we have two ball. One is moving with velocity (0, 4 m/s), while the other is moving with velocity (3 m/s, 4 m/s). The magnitude of the drag force experienced by a sphere is given by
F = 1/2 rho C[sub]D[/sub] A u^2,
where rho, C_D, and A are constants and u is the velocity. For the first ball, the magnitude of the drag force is
Another way of looking at this is to see where the v^2 comes from in the first place. One factor of v is because the faster the air molecules are moving relative to the ball, the more momentum they’ll impart to the ball during the collision. This factor of v is only the vertical component of v. But the other factor of v is because the faster you’re moving, the more air molecules you’ll collide with in any given amount of time. And for this factor of v, it’s the total motion that’s relevant.
(I’m starting to feel like Indistinguishable here, with three posts in a row which each elaborate on the previous one)
Oh, and if anyone’s keeping track, MikeS and I are saying the same thing. My constant c is the same as his 1/2 * rho * C[sub]D[/sub] * A. But since those are all constants, and the equations are long enough and annoying enough to type out as it is, I lumped them all into a single constant.
Chronos, suppose you started this experiment 6" off the ground and worked upward 1 foot at a time. Instead of throwing the ball you had a loaded piston mounted horizontal that would poke the ball dead center a wide range of speeds. Would the power of the poke or the height of release make any difference as to which landed first?
IANAP, but I just now made a simulation to see what happens. Granted, this is a discrete approximation to a continuous case, but it should give us some insight.
I made a spreadsheet to calculate position x, position y, velocity x, velocity y, combined speed, drag, then drag x and drag y. I used an adjustable drag factor and time increment dt. For each line of the spreadsheet, it updates pos x and pos y based on the previous line’s vel x and vel x, respectively, times dt. Then I calculated combined speed s = sqrt(Vx^2+Vy^2), multiplied by the drag factor to get drag, then broke that down into drag x and drag y based on Vx and Vy. Because of similar triangles, drag x equals drag times Vx over s and drag y equals drag times Vy over s. Then I updated Vx as the previous Vx minus drag x times dt, and Vy as previous Vy plus (9.8 minus drag x) times dt.
I set the drag factor=.001 times the speed squared, which gives terminal velocity of just under 100 m/s, about 221 mph (which seems reasonable to me). Then I lowered dt to 0.1 to see what happens after falling for just 4.0 seconds. An object with initial Vx=0 fell 74.656 meters but an object with initial Vx=10 m/s fell only 74.466 meters. Just for fun, I changed the drag formula to be linear instead of quadratic and found that the two objects fell exactly the same distance. This verifies Chronos’s answer and reassured me that my simulation is working. An object thrown sideways does indeed fall slower, but the difference is small. Putting the drag formula back the way it was and making initial Vx=47 m/s (the fastest baseball pitch ever recorded) increased the effect. Now the thrown object only fell 73.36 meters in 4 seconds. With Vx=100 (thrown by a robot arm, I guess?), it fell 72.43 meters in 4 seconds. With initial Vx=200 m/s, it only fell 71.619 meters. That’s still a small difference.
Bottom line, an object thrown sideways off a 75 meter cliff (or a 25-story building) will fall about 1% slower than the same object dropped straight down.
For those of us who slide through Physics 101 with a C- largely because their professor never wanted to see them again next semester, is there an answer here? In English?
In all cases, which one lands first would always be the same: The one that dropped straight down. How great the difference is in landing times would depend on the other variables, but there’s no way for those conditions to lead to the “poked” one landing first.
Throw it hard enough, or from high enough, and the thrown ball will have further to fall as well, due to the curvature of the Earth. If you can throw it so that it lands a mile away from your starting point, it has to fall about 8 more inches than the ball dropped straight down.
Here’s the simplest I can make it. Drop a baseball off the roof of a hotel, taking video with your cell phone. It will gain speed on its way down but will lose some of that speed because of drag (wind resistance). Throw another baseball horizontally off the roof of the hotel, and take video of that one two. Afterwards, let’s freeze frame the second video about one second in, at a point in time where the thrown baseball is going about 40 mph at a 30 degree downward angle. The ball is moving both horizontally and vertically. It’s moving downward at about 20 mph and horizontally at about 35 mph. Now compare this to a similar point in the first video, when the first baseball is moving straight down at about 20 mph. The thrown baseball has TWICE as much speed as the dropped one. But twice as much speed means FOUR TIMES as much drag. Most of the drag on the thrown baseball is horizontal, but just the vertical portion is twice as much as the drag on the dropped baseball. The thrown ball will lose some of its horizontal speed and also lose some of its vertical speed. The amount it loses vertically will be TWICE as much as the dropped ball is losing at that point. The dropped baseball loses maybe 1% of its vertical speed due to wind resistance but the thrown baseball loses 2%. So the dropped baseball lands sooner, by about a 1% difference.
The dropped bullet will hit the ground before the bullet fired from the gun. The bullet fired from the gun has a nose high trajectory which (along with drag) creates aerodynamic lift.