Photo by Jeff Ott
One of the first topics in a physics course is the motion of
projectiles, such as a shotput. A shotput is a very ideal projectile
to look at because it isn't affected much by the air it moves through.
One can therefore ignore effects like air drag or aerodynamic lift in
the case of a shotput.
You have perhaps heard that a shotput follows a parabolic
trajectory. A parabola is a line that is given by the equation
y(x) = a + bx + cx²,
where a,b,c are constants. Why does a shotput follow a parabolic
arc, and not some other curve?
The answer lies in the nature of motion in the presence of gravity.
And to look at that, let's first drop the shotput from the top of a
wall of height h (making sure no one we like is standing below). If
we record its fall with a video or film
camera, we can write its height y as a function of time t. Say
the wall has a height of 10 m (a very high wall); you'd get a table
of y(t) such as second column below (units are put in brackets,
like [s] for seconds and [m] for meters):
| t [s] || y [m] || v [m/s] |
| 0.0 || 10.00 || 0.00|
| 0.1 || 9.95 || -0.98|
| 0.2 || 9.80 || -1.96|
| 0.3 || 9.56 || -2.94|
| 0.4 || 9.22 || -3.92|
| 0.5 || 8.77 || -4.90|
| 0.6 || 8.24 || -5.88|
| 0.7 || 7.60 || -6.86 |
| 0.8 || 6.86 || -7.84 |
| 0.9 || 6.03 || -8.82 |
| 1.0 || 5.10 || -9.80 |
| 1.1 || 4.07 || -10.78 |
| 1.2 || 2.94 || -11.76 |
| 1.3 || 1.72 || -12.74 |
| 1.4 || 0.40 || -13.72 |
It takes a little over 1.4 s for the shotput to drop from a height of
Note that the shotput falls larger distances in equal time
increments as time increases; in other words, its speed, the
rate at which it drops, increases with time.
Suppose in addition to
the film camera you had one of those guns the cops use to measure
speed (maybe I'll show how those work in a later chapter).
Then you'd get the speed measurements v(t) shown in the third
Notice that the downward (negative) speed increases linearly: each
0.1 s interval adds -0.98 m/s to the speed. This is known as
constant acceleration: the acceleration of an object due to
gravity near the surface of the Earth is a constant 0.98 m/s per 0.1
s, or 9.8 m/s per s downward, written 9.8 m/s². This constant is usually
called g, and it's an important number for projectile motion:
g = 9.8 m/s² = 32 ft/s².
If, instead of dropping the shotput from rest at the top of the wall,
we threw it upward, then it would start with a positive vertical
speed, travel upward, reach a maximum height and then fall back
down. There is really no difference in this case, and the vertical
speed still has -9.8 m/s added to it every second by gravity. This is
exactly what happens with the shotput!
While a shotput is traveling horizontally downfield, it is actually
undergoing the exact same vertical motion that it would have if just
thrown straight up.
The shotput toss is just a combination of constant horizontal motion
and downward vertical acceleration. Here's what it looks like from the side (hit
RELOAD to replay this GIF animation 10 times):
I've used numbers approximately matching the world record
shotput trajectory of Randy Barnes, USA, 1990. I've assumed he
launched the shotput from a height of 2 m at a 45-degree angle to the
ground. I label the vertical position of the shotput
with y and the horizontal position (distance downfield) with
Since the shotput travels an equal horizontal distance every 0.1 s,
the horizontal velocity vx is constant!
Gravity has no effect on horizontal motion, so the shotput
simply coasts with the initial horizontal velocity that Randy gave it.
However, gravity does affect the vertical velocity
vy in the same way it did in the earlier example:
vy changes by -0.98 m/s in each 0.1 s interval.
The combination of constant horizontal velocity and constant downward
acceleration results in a parabolic trajectory as seen above. You can vary the launch angle and initial speed and see the resulting trajectories with the Projectile Applet.
Here are the equations that govern projectile motion, in case you need
them. The initial conditions are that the object starts at
x0 and y0 with horizontal and vertical
speeds vx0 and vy0.
- horizontal velocity: vx = vx0
- vertical velocity: vy = vy0 - gt
- horizontal position: x = x0 + vx0t
- vertical position: y = y0 + vy0t -
If one eliminates t from the last two equations, one gets the
parabolic equation for y(x).
- Gravity near the Earth's surface accelerates an object downward
with the constant value g = 9.8 m/s².
- Projectile motion is the combination of constant
horizontal motion and downward acceleration due to gravity.
- The resulting trajectory is a parabola.