# Projectile motion

Projectiles are objects which have a velocity but no engine providing thrust. The study of projectile motion was very important in the history of physics, since it provided Galileo and Newton with the observations and measurements which led to an understanding of gravity and motion, and from these force and energy.

Let us start our investigation with a simple type of projectile motion: the horizontal launch:

## Horizontal Launch

#### a simple experiment

Place two identical coins on the edge of a table. Try flicking one off the table so it flies horizontally, and at the same time nudge the other coin off the edge. It takes a bit of practice to get them to leave the table's support simultaneously.

Which hits the floor first?

Now try the same thing, but with coins of different sizes and masses.

What do you observe?

Projectiles on Earth with horizontal constant velocity will accelerate vertically due to gravity. The resulting flightpath, or trajectory, is parabolic.

Since the horizontal velocity is constant, the horizontal displacement is:

\$\$x = v_0t\$\$

where \$x\$ is the displacement along the x-axis, after time \$t\$, when travelling at constant horizontal velocity \$v_0\$, which is also the initial velocity of the object, since there is zero vertical velocity initially.

In the vertical direction, the velocity is given by: \$v_y = -gt\$

Taking the integral, we find that the vertical displacement can be expressed as:

\$\$y = -1/2gt^2\$\$

We can combine the two equations to express the vertical displacement in terms of x and the initial velocity, without the time:

\$\$y = -1/2g(x/{v_0})^2\$\$
Constant horizontal velocity and constant vertical acceleration produce a curve in the shape of a parabola.

#### Plotting Flight

Calculate how far an object falls after 0.1s, 0.2s, 0.3s, etc.

Cut strings to the lengths you calculated, and tie weights on the ends. Hang these at equal intervals on a wall or board.

Throw a ball so that it has a horizontal speed. Can you get the ball's path to match the shape of the path marked out by the rings? What is this shape?

## Launch at an Angle

Of course, projectiles can also be launched at angles to the horizontal. Cannons fire cannonballs, and footballers kick balls, at an angle to the horizontal, to gain more height, or increase the horizontal distance the projectile travels (range).

If θ is the angle of launch to the horizontal, the initial horizontal velocity is: \$v_x = v_0⋅cosθ\$

Since \$v = x/t\$, and the velocity is constant:

\$\$x = v_0tcosθ\$\$

The vertical motion is more complicated because there is constant acceleration in the opposite direction to the initial velocity \$v_0sinθ\$.

The vertical velocity is: \$v_y = v_0sinθ - gt\$, which gives us a displacement of:

\$\$y = v_0tsinθ - 1/2gt^2\$\$

### Apogee

Often, it is useful to be able to calculate the apogee, whether as the maximum height reached, or the time it takes to reach the highest point. This can be found by realising that the maximum height is when \$v_y = 0\$, since the projectile is about to change vertical direction and start falling.

\$0 = v_0sinθ - gt\$, so \$t_{apo} = {v_0sinθ}/g\$, where \$t_{apo}\$ is the time from launch to the apogee.

To find the highest point in the trajectory, we need to find \$y_{max}\$ at time \$t_{apo}\$:

\$y_{max} = v_0{v_0sinθ}/g{sinθ} - 1/2g({v_0sinθ}/g{sinθ})^2 =\$

\$\$y_{max} = {v_0^2sin^2θ}/{2g}\$\$

### Range

The range is the maximum horizontal distance travelled by the projectile. Assuming the ground is level for the whole flight, the projectile hits the ground when \$y = 0\$:

\$0 = v_0tsinθ - 1/2gt^2 = t(v_0sinθ - 1/2gt)\$

#### Maths check

\$\$sin2θ = 2sinθcosθ = 2sinθsin(90 - θ) \$\$ \$\$sin2(90 - θ) = 2sin(90 - θ)sinθ \$\$

This gives two solutions. \$t = 0\$ (at launch) and \$t = {2v_0sinθ}/g\$.

The range is the displacement:

\$x_{max} = v_0tcosθ = {2v_0^2sinθcosθ}/g \$, or:

\$\$⇒ x_{max} = {v_0^2sin2θ}/g\$\$

\$sin2θ\$ has a useful property: \$sin2θ = sin2(90 - θ)\$

Therefore, if the two angles are \$θ\$ and \$90 - θ\$, or in other words the two angles sum to 90°, then the range will be the same!

For example, a projectile launched at 30° will travel just as far as a projectile launched at 60°.

### Flightpath Shape

It is common knowledge that a projectile follows a parabolic path, if air resistance is negligible. But how can this be proven mathematically?

\$x = v_0tcosθ\$, or \$t = x/{v_0cosθ}\$

\$y = v_0tsinθ - 1/2gt^2 = v_0sinθ(x/{v_0cosθ}) - 1/2g(x/{v_0cosθ})^2 \$

\$= xtanθ - g/{2v_0^2cos^2θ}x^2\$

which is an equation which describes a parabola.

### Conservation of Energy

A projectile at launch has a total energy equal to its kinetic energy: \$E = 1/2mv_0^2\$.

When it is at height \$h\$, it has a total energy of \$E = 1/2mv^2 + mgh\$.

Equating the two equations: \$E = 1/2mv_0^2 = 1/2mv^2 + mgh\$

\$v^2 = v_0^2 - 2gh\$

### Air Resistance

The equations for projectile motion assume zero or negligible air resistance. That may be true for playing golf on the Moon, but on Earth it is never the case.

Air resistance increases with the speed of the moving body, its cross-section area (the surface exposed to the 'wind'), and decreases with increasing density of the body.

Since the projectile slows down as it rises, the air resistance decreases, and increases again as it falls. This results in a trajectory which is not parabolic, and not symmetrical. Compared to a projectile with no air resistance, the projectile travels less distance, less height, and impacts the ground at a steeper angle than the launch.

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