What is 2D projectile motion?

In a fructose induced rage, you decide to throw a lime at an angle through the air. It takes a path through space as shown by the curved, dashed line in the diagram below. The lime in this case is considered to be a two-dimensional projectile since it's flying both vertically and horizontally through the air, and it's only under the influence of gravity.

Since the gravitational force pulls downward, gravity will only affect the vertical component of the velocity $\red {v_y}$start color #df0030, v, start subscript, y, end subscript, end color #df0030 of the lime. The horizontal component of the velocity $\blue {v_x}$start color #6495ed, v, start subscript, x, end subscript, end color #6495ed will remain unaffected and stay constant as the lime moves along its path.

Try sliding the dot in the diagram below to see that the vertical velocity $\red {v_y}$start color #df0030, v, start subscript, y, end subscript, end color #df0030 changes, but the horizontal velocity $\blue {v_x}$start color #6495ed, v, start subscript, x, end subscript, end color #6495ed remains constant.

Concept check: At the maximum height of the lime's trajectory, what is the value of the vertical component of velocity?

One of the easiest ways to deal with 2D projectile motion is to just analyze the motion in each direction separately. In other words, we will use one set of equations to describe the horizontal motion of the lime, and another set of equations to describe the vertical motion of the lime. This turns a single difficult 2D problem into two simpler 1D problems. We're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime. Similarly, throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime. In other words, if you fire a bullet horizontally and drop a bullet at the same time, they will hit the ground at the same time.

There's no acceleration in the horizontal direction since gravity does not pull projectiles sideways, only downward. Air resistance would cause a horizontal acceleration, slowing the horizontal motion, but since we're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile.

So for the horizontal direction we can use the following equation,

Note: Be sure to only plug horizontal variables into this horizontal equation. If we know two of the variables in this equation we can solve for the remaining unknown variable.

Two-dimensional projectiles experience a constant downward acceleration due to gravity $a_y=-9.8 \dfrac{\text{m}}{\text{s}^2}$a, start subscript, y, end subscript, equals, minus, 9, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction. Since the vertical acceleration is constant, we can solve for a vertical variable with one of the four kinematic formulas which are shown below.

Be sure to only plug vertical variables into these vertical equations. If we know three of the variables in these equations we can solve for any of the remaining unknown variables.

Note: For a given process, the time interval $t$t has the same value for the vertical and horizontal equations. This means that if we ever solve for the time $t$t, we can plug that time $t$t into the equations for either the vertical or horizontal directions. This strategy is used in many problems. Often, one has to solve for time $t$t using the vertical equations, then plug that time into the horizontal equation (or vice versa).

Many times, people try to substitute vertical components into a horizontal equation, or vice versa. Analyzing each direction (horizontal and vertical) of a projectile independently only works if you keep the different directions ($x$x or $y$y) in their own separate equations.

Initial velocities that are directed diagonally will have to be broken into vertical and horizontal components. People sometimes have a hard time breaking a velocity vector into vertical and horizontal components. See this article for help with the trigonometry you use to break vectors into components.

When a projectile is shot horizontally, the initial vertical velocity is zero $\red {v_{0y}=0}$start color #df0030, v, start subscript, 0, y, end subscript, equals, 0, end color #df0030 (see example 1 below). Many learners have a hard time understanding that an object can start with a horizontal component of velocity, yet have zero vertical component of velocity.

A water balloon is thrown horizontally with a speed of $v_0=8.31 \dfrac{\text m}{\text{s}}$v, start subscript, 0, end subscript, equals, 8, point, 31, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction from the roof of a building of height $H=23.0\text{ m}$H, equals, 23, point, 0, start text, space, m, end text.

We can start by drawing a diagram that includes the given variables.

Once we find the time of flight $t$t, we'll be able to solve for the horizontal displacement using $\Delta x=v_xt$delta, x, equals, v, start subscript, x, end subscript, t. To solve for time, consider the fact that we know three variables in the vertical direction ($\Delta y=-23.0\text { m}$delta, y, equals, minus, 23, point, 0, start text, space, m, end text, $v_{0y}=0$v, start subscript, 0, y, end subscript, equals, 0, $a=-9.8\dfrac{\text m}{\text {s}^2}$a, equals, minus, 9, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction).

So we'll use a kinematic formula in the vertical direction to solve for time $t$t. We don't know the final velocity $v_y$v, start subscript, y, end subscript, and we aren't asked for the final velocity $v_y$v, start subscript, y, end subscript so we'll use the vertical kinematic formula that doesn't include final velocity.

Now we need to plug this time $t$t into the equation for the horizontal direction to find horizontal displacement $\Delta x$delta, x.

$\Delta x=(8.31 \dfrac{\text m}{\text s})(2.17\text { s}) \quad \text{(plug in the time of flight and } v_x)$delta, x, equals, left parenthesis, 8, point, 31, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction, right parenthesis, left parenthesis, 2, point, 17, start text, space, s, end text, right parenthesis, start text, left parenthesis, p, l, u, g, space, i, n, space, t, h, e, space, t, i, m, e, space, o, f, space, f, l, i, g, h, t, space, a, n, d, space, end text, v, start subscript, x, end subscript, right parenthesis

So the water balloon struck the ground $18.0 \text{ m}$18, point, 0, start text, space, m, end text horizontally from the edge of the building.

An air cannon is used to launch a pumpkin off a cliff of height $H=18.0 \text{ m}$H, equals, 18, point, 0, start text, space, m, end text with an initial speed $v_0=11.4 \dfrac{\text m}{\text s}$v, start subscript, 0, end subscript, equals, 11, point, 4, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction at an angle of $\theta=52.1^\circ$theta, equals, 52, point, 1, degrees as seen in the diagram below.

What is the speed of the pumpkin right before it hits the ground?

We'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity ($v_x$v, start subscript, x, end subscript and $v_y$v, start subscript, y, end subscript).

Before we can do that, we'll have to find the components of the initial velocity ($v_{0x}$v, start subscript, 0, x, end subscript and $v_{0y}$v, start subscript, 0, y, end subscript) using the definitions of sine and cosine.

(Note: If that seemed like indecipherable mathematical witchcraft, check out this article for help with breaking vectors into components.)

This value we found for the horizontal component of the initial velocity $v_{0x}=7.00 \dfrac{\text m}{\text s}$v, start subscript, 0, x, end subscript, equals, 7, point, 00, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction will also be the horizontal component of the final velocity $v_x=7.00 \dfrac{\text m}{\text s}$v, start subscript, x, end subscript, equals, 7, point, 00, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction since the horizontal component of the velocity remains constant for the entire flight (assuming there is no air resistance).

To find the vertical component of the initial velocity we'll use the same procedure as above but with sine instead of cosine.

Since the vertical component of velocity $v_y$v, start subscript, y, end subscript changes for a projectile as it moves through the air we'll have to solve for the vertical component of the final velocity $v_y$v, start subscript, y, end subscript using a kinematic formula for the vertical direction. Since we don't know the time of flight $t$t, and we weren't asked to find the time $t$t, we'll use the vertical kinematic formula that doesn't include time $t$t.

Now that we know the horizontal and vertical components of the final velocity, we can use the Pythagorean theorem to solve for the final speed (i.e. magnitude of the final velocity).

This speed of $v=21.9 \dfrac{\text {m}}{\text {s}}$v, equals, 21, point, 9, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction is the magnitude of the final velocity of the pumpkin right before it hits the ground. The relationship between the final velocity and its components is shown in the diagram below.

We could also solve for the angle $\phi$\phi of the final velocity using the definition of tangent.

$\text{tan}\phi=\dfrac{20.8\dfrac{\text m}{\text s}}{7.00\dfrac{\text m}{\text s}}$start text, t, a, n, end text, \phi, equals, start fraction, 20, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction, divided by, 7, point, 00, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction, end fraction

Now taking inverse tangent of both sides we get,

The left hand side just becomes $\phi$\phi, and we can find the value of the right hand side by plugging into a calculator to get,