Motion in a plane and projectile motion

Two axes describe a position

In a plane we need two coordinates: r = (x,y). Choose an origin, a horizontal x axis and a vertical y axis positive upwards. Position can be described by an arrow from the origin to the body; displacement between instants has components Δx and Δy. Its magnitude is √[(Δx)² + (Δy)²].

Velocity also has components vₓ and vᵧ. Speed is v = √(vₓ² + vᵧ²). A zero component does not require a zero vector: at the highest point of an oblique launch, vᵧ = 0 but vₓ is generally not zero. Velocity direction is tangent to the trajectory.

If launch speed is v₀ and angle θ is above the horizontal, v₀ₓ = v₀ cos θ and v₀ᵧ = v₀ sin θ. These results come from projecting the right triangle formed by the vector and its components. Components retain units of m/s; sine and cosine have no unit. Check whether the calculator uses degrees or radians according to the angle given.

What makes a body a projectile in the model

After launch and before impact, we consider weight to be the only significant force. We neglect air resistance, propulsion and rotational effects. Near the surface, the gravitational field is approximately uniform. We will use g = 9,8 m/s² in all examples in this lesson.

By the second law, aₓ = 0 and aᵧ = −g. Thus horizontal velocity stays constant and vertical velocity varies uniformly. Independence means that vertical force does not directly change the horizontal component. It does not mean there are two bodies or two clocks: components describe the same body over the same time.

A dropped ball and another launched horizontally from the same height reach the ground simultaneously in this model if they start with the same vertical velocity. The second travels farther horizontally, but that advance does not delay falling. In the real world, air, shape and spin can change the result, especially for light objects or long trajectories.

Building the equations by component

If t = 0 is the launch instant, the horizontal equations are x = x₀ + v₀ₓt and vₓ = v₀ₓ. Since aₓ = 0, there is no quadratic term in x. Horizontal distance only reveals time if v₀ₓ is non-zero and the motion follows this model.

Vertically, we apply constant acceleration: y = y₀ + v₀ᵧt − gt²/2 and vᵧ = v₀ᵧ − gt. Also, vᵧ² = v₀ᵧ² − 2g(y − y₀). This last relationship avoids time but loses the sign of vᵧ through squaring; rising or falling must be identified physically.

To find the instant of an event, impose its condition: the top means vᵧ = 0; ground means y equals ground height; passing a wall means x equals the wall’s position. Solve for time first and then use it in the other component. Reject negative roots when the interval studied begins at launch.

Why the trajectory is parabolic

When v₀ₓ ≠ 0, the horizontal relationship gives t = (x − x₀)/v₀ₓ. Substituting into the vertical equation gives y − y₀ = (v₀ᵧ/v₀ₓ)(x − x₀) − [g/(2v₀ₓ²)](x − x₀)². This is a quadratic function of the horizontal coordinate: the trajectory is a parabola in the adopted model.

It is concave down because the quadratic coefficient is negative. The curve drawn in space is not the same as a y(t) graph, although both are parabolic here. One uses horizontal distance on its axis; the other uses time. Axes and units tell an essential part of the story.

For an exactly vertical launch, v₀ₓ = 0 and the previous elimination involves division by zero: it does not apply. The trajectory is a vertical line, described only by y(t). A formula has a domain of validity; recognising a special case is part of solving the problem, not an optional detail.

The top and return to the initial height

At the top, 0 = v₀ᵧ − gt, so top t = v₀ᵧ/g for an initially rising launch. Substituting into position gives the height gained: maximum Δh = v₀ᵧ²/(2g). This is the increase relative to launch; height above ground still includes y₀.

If arrival is at exactly the initial height, the vertical equation gives t(v₀ᵧ − gt/2) = 0. Besides the launch root t = 0, flight time is 2v₀ᵧ/g. Horizontal range is R = v₀ₓt = v₀² sin(2θ)/g. The symmetry of ascent and descent depends on returning to the same level and retaining the model’s assumptions.

For fixed initial speed and arrival at the same height, sin(2θ) is maximised at θ = 45°. Complementary angles give the same ideal range but different heights and times. If arrival is above or below launch, the previous R expression and the 45° statement are no longer general. Solve the equation using the actual height difference.

A solution should end with interpretation

Organise data into two groups of components, but use a common timeline. Write initial and final conditions with signs; for example, a point 5 m below the origin has y = −5 m. Do not always substitute a positive number for “height” without choosing the reference frame.

After finding vₓ and vᵧ at an instant, calculate magnitude and direction if the question asks for complete velocity. The angle can be obtained from tan φ = vᵧ/vₓ, but the quadrant depends on the signs. At impact after falling, vᵧ must be negative with an upward axis; a positive square root may indicate only magnitude.

Check the results: t must be positive for future flight; range must have the direction of v₀ₓ; at the top, speed is |v₀ₓ|, not zero. An energy comparison, v² = v₀² − 2g(y − y₀), can check the final magnitude when only gravity does work.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Now try it yourself

There is no timer. Write what you know, choose a relationship and justify each step. Hints and solutions are available whenever you need them.

Mistakes worth avoiding

At the top the entire velocity is zero.

Only the vertical component is zero in a usual oblique launch.

Horizontal motion delays the fall.

Under the model’s assumptions, vertical evolution depends only on vertical conditions.

45° always gives the greatest range.

The conclusion requires equal launch and arrival heights and the model without air.