Compare this behaviour of a hammer and feather dropped on Earth (in the atmosphere) with a similar experiment peformed on the moon
(in near vacuum) by Commander David Scott of the Apollo XV mission. To see the complete NASA movie, follow this link. For more historic NASA footage, try this link.
Nevertheless, objects often accelerate towards the Earth with very nearly the same acceleration, as do the apple and the grape in this example. In this page on projectiles, we look at this behaviour and some of its consequences.
We return to look quantitatively at the effects of the earth's atmosphere below. Briefly, however, the air resistance will be small compared to the weight if:
a) the weight is large,
b) the object has a small cross-section normal to the direction of motion,
c) the object is streamlined, and
d) the speed is low.
The grape rolls off a table, 77 cm high. Because it rolls horizontally on the table, its initial vertical component of
velocity is zero. Using the step button, we observe that the grape took 10 frames to travel 56 cm. The camera runs at 25 frames per second, so the horizontal component of velocity is 1.4 m.s-1, in agreement with the calculation. Note the angle of the rebound: because the grape rolled off the table, it was spinning when it hit the carpet. This caused it to accelerate to the right during the collision.
If we neglect air resistance, then the acceleration is − g downwards: there is no horizontal acceleration. As we have seen above, horizontal and vertical motion are independent. So we can write the equations for motion in the vertical and horizontal motion as we did in the introductory module:
The second of these equations -- curently giving x as a function of t -- may be rearranged to give t as a function of x. Doing so, and substituting for t in the first equation gives:
This equation, which expresses y in terms of x, is the equation of the trajectory of the projectile. (It looks less formidable -- see below -- if we choose the position of the origin so that x0 and y0 are zero.)
Suppose that you launch a projectile at 20 m/s, and that the launch speed is independent of the launch angle. Further, let's launch it from ground level.
How far will it go, and what angle gives the greatest range?
We choose the origin to be the position from which we are launching. So x0 and y0 are zero. The equation above now simplifies to:
In this case, our landing site is at y = 0. The range R is the x position at which it lands -- in other words, it is the value of x for which y is zero. In yet other words, the position (R,0) is the position at which the parabola intersects the x axis.
Note, by the way, that the parabola is dotted below the x axis. This part of the curve is given by the same equation for the motion of a projectile, but below the axis our object is no longer a projectile: it is a buried, rather than being in free fall. It is rather common in physics that equations produced by applying a general principle to a particular case have only a limited range of applicability.
So, if we set x = R and y = 0 in the equation above, we have:
We want to know R. So we rearrange the equation above to give
In this problem (as is often the case) we interesting properties of the launch velocity are its magnitude and direction. So we must express components vx and vy0 in terms of v0 and θ. Simple trigonometry gives us the first equation below, and the second uses the trigonometric identity that sin 2θ = 2 sin θ.cos θ:
The maximum in this function of θ occurs at the value of &theat; at which the function is not increasing or decreasing as a function of &theata, ie when its derivative is zero. Let's take the derivative, remembering to use the chain rule. (If necessary, see A simple but useful introduction to calculus.) This gives:
So the maximum range occurs when cos 2θ is zero. What angle is this? In the animation below, the launch angles are 15°, 30°, 45°, 60°, 75° and 90°. The animation is shown at two thirds of real speed.
So, should you throw a ball at 45° to achieve maximum range? No. First, we assumed above that launch speed was independent of angle. This is not the case for throwing: most people can throw considerably faster at low angles. There's one reason for throwing at a flatter angle.
Second, we have neglected air resistance. Although this is a small effect for low speeds and large objects, it is often non-negligible for balls, as we show below. For objects that, like balls, are symmetrical about the line of their velocity, the resistance always opposes the motion, so it means that the vertical component of velocity decreases more rapidly during ascent, and increases less rapidly during descent, than would be the case in a vacuum. Throughout the whole trajectory, it reduces the horizontal component of velocity. The consequences of this is that the ball falls short of the trajectory calculated by neglecting air resistance. For this reason too you should throw with θ less than 45°. (For a frisbee, a discus or a javelin, whose axes of symmetry are not parallel to the velocity, aerodynamic lift is often important.)
This non-streamlined bus, with cross-sectional area A, is travelling at velocity v. What force is exerted by the air on the bus?
When an object moves through a fluid at sufficiently low speed, the drag force is due to viscosity, which gives rise to a force somewhat like friction. (Without going into viscosity here, honey has greater viscosity than water, which has greater viscosity than air.) At sufficiently high speeds, the drag is due to the acceleration of the nearby fluid, and the subsequent loss of kinetic energy by that fluid in turbulent flow. For macroscopic objects at ordinary speeds in air, the drag is almost entirely turbulent, and that is what we shall analyse.
In a time t, the bus travels vt. In that time, approximately all of the air in a volume A.vt must be accelerated to the speed of the bus. The density of the air is ρ, so the mass of this volume is ρ.Avt. So the air will gain a kinetic energy:
ΔK = (1/2)mv2 ≈ (1/2)ρ.Avt.v2
The work to produce this kinetic energy is done by the bus, which pushes the air forwards. From Newton's third law, the force that the bus exerts on the air equals in magnitude that which the air exerts on the bus. The latter force is called the drag, Force, Fdrag. So the work done to accelerate the air is F.vt.
From the work-energy theorem, W = ΔK, so in this case
Fdrag.(vt) = ΔK ≈ (1/2)ρ.(Avt).v2.
Cancelling the vt gives:
Fdrag ≈ (1/2)A.ρ.v2.
The "≈" came because we said that approximately all of the air in front of the bus is accelerated to match its speed. This is a bit complicated, because the air is pushed sideways, and some of it is accelerated to speeds less than v. Some air is drawn behind the bus, too. Further, it depends on how streamlined the object is, on both front and back.
There is no simple way to calculate the drag force. For a given object, it can be measured. From such a measurement, we define the drag coefficient, CD thus:
CD depends strongly on the geometry, It depends much less strongly on the size and speed of the objects. So one can measure CD for an object in a wind tunnel, and use that value to estimate Fdrag for other objects with similar shapes. For a sphere at ordinary speeds, the value is about 0.5. For streamlined automobiles it is about 0.3-0.4. For a cyclist it is a bit less than 1.
In the multimedia module, we referred to the force on a hand held out of a moving car. Taking just order-of-magnitude values, let's put 100 km/hr, CD ~ 1, and A ~ 0.01 m2. The density of air is 1.2 kg.m−3. Substituting in the equation above, the force is several newtons.
When can one ignore air resistance?
For an object with circular cross section, let's define the fraction:
For a grape, m is about 5 g and r about 10 mm. For an apple, m is about 100 g and r about 30 mm. For a sphere, the drag coefficient is about 0.5. The density of air is 1.2 kg.m−3. So, at a speed of 5 m.s−1,f is approximately 0.05 for the grape and 0.02 for the apple. Dropped from the hand, the objects don't go much faster than this, and are slower for much of the trajectory. So the difference is small.
Balls, however, are thrown, hit and kicked much faster. Further, while some balls (cricket, golf) are dense, others (footballs) have low density. How important is drag? Why not do some examples?
A projectile question: When does the cricket ball travel its fastest (towards the batsman)?
This question was asked by Dan Trouw of Darwin. Here is the situation. The ball is released from the hand with the arm outstretched above the bowler's head, so it typically has a little more than two metres to fall to ground level. On the other hand, air resistance is slowing it from the moment it leaves the hand. Which force 'wins'?
The answer is not immediately obvious -- to me, at least. Although a cricket ball is dense, from observatoin, air resistance is not negligible: The trajectory of a well-hit ball is noticeably not a parabola: it falls more rapidly than one would expect. (My excuse for missing that important catch on the boundary, and I am sticking to it.).
As above,
Fdrag = (1/2)CDAρv2,
For a sphere CD is typically about 0.5. So we have
Fdrag ≈ 0.25πv2ρv2,
For a very fast bowler, with v = 40 m/s, this force is about 2 newtons. For such a bowler, the drag force is about the same as the weight of the cricket ball. For someone bowling at say 20 m/s, the drag force would be only a few tenths of the weight.
Thus, for a ball bowled vertically downwards (a new meaning for shooting one's self in the foot), and for a slow bowler, weight would be more important than drag, as it clearly is when the ball is dropped.
However, the ball is not bowled vertically down -- at least not deliberately -- and that is important for this question. The effect of direction can be included by considering kinetic energy. Gravity has only about two metres of downwards vertical travel over which to accelerate the ball, while the drag has about twelve metres (more for a slow bowler) to decelerate it. And a very slow bowler would have to bowl the ball slightly upwards ("give it a bit of air") just so that it covers the distance. In which case, gravity would act to slow it at first, not accelerate it.
Consequently, drag has a greater effect on reducing the kinetic energy (and thus slowing it) than has gravity in increasing it. So, even for quite slow bowlers, the fastest that the ball is travelling is just when it leaves the bowler's hand -- until, of course, it reaches the bat.
