Horizontal projectile motion

With vertical projectile motion, we only had to worry about one component: the vertical component.

With horizontal projectile motion though, gravity still applies, so we have to work with two different components of the motion: the horizontal component and the vertical component.

Horizontal component

Velocity

Gravity works on the vertical component of the motion (see vertical projectile motion), so weight has no effect on the horizontal component of the motion.

There’s no other forces involved after the projectile is launched! That means that the horizontal velocity of the projectile will stay constant throughout the motion.

Horizontal velocity is constant.

The horizontal velocity will be the same as it was launched with: the initial horizontal velocity.

Displacement

Because \text{displacement} = \text{velocity} \times \text{time}, the horizontal displacement of the projectile will be the horizontal velocity multiplied by the time it’s in the air for.

Acceleration

There’s no forces acting on the horizontal component of the motion, so the acceleration of the projectile in the horizontal direction will be zero.

Vertical component

The vertical component of the motion is the same as it is in vertical projectile motion, so we can use the same equations and values as in that chapter to solve problems involving the vertical component of the motion. This isn’t the same thing though, as the object isn’t being launched upwards vertically first.

Acceleration

The acceleration which acts vertically is caused by gravity, so the acceleration of the projectile in the vertical direction will be -9.81ms^{-2} (negative because it’s downwards).

Vertical acceleration is -9.81ms^{-2}.

Velocity

The velocity will start at 0ms^{-1} (since it’s launched horizontally, so no initial vertical velocity), and will increase in the negative direction (downwards) as the projectile falls, due to the acceleration of gravity.

We can use \text{velocity} = \text{acceleration} \times \text{time} to find the vertical velocity at any point in time while the projectile is falling.

\text{Vertical velocity} = -9.81ms^{-2} \times \text{time}

Displacement

We can use \text{displacement} = \frac12 \times \text{acceleration} \times \text{time}^2 to find the vertical displacement of the projectile at any point in time while it’s falling.

We can find that equation from the SUVAT equation s=ut+\frac12at^2 by substituting u=0ms^{-1} (since the projectile is launched horizontally, so no initial vertical velocity).

Time

The time it’s in the air for can be found using any of the equations for uniform acceleration.

We can also just use the standard equations for displacement, velocity and acceleration to find the time, since we may know the values for those variables.

Time is shared

The only value which is guaranteed to be the same for both the horizontal and vertical components of the motion is the time. The time it’s in the air for is the same for both components, because the projectile lands at the same time in both the horizontal and vertical directions.

That’s really useful to know, because it means that if we know the time for one of the components, we can use that time to find the other values for the other component.

Using the right values

When you have a question like this: you will have two values for certain variables. For example, you will have two values for velocity: the horizontal velocity and the vertical velocity, as well as all the other variables.

You MUST make sure that you use the right values for the right component of the motion. The horizontal displacement, for example, has absolutely nothing to do with the vertical displacement.

This applies for all the variables, except for time, which is shared.

Drawing of horizontal projectile motion

If we trace the path the projectile takes, it’s a curve which slopes downwards.

That’s because, as time goes on, the vertical velocity increases in the negative direction (downwards) due to the acceleration of gravity, while the horizontal velocity stays constant. So the gradient of the curve gets steeper and steeper as the projectile falls, because the vertical velocity is increasing (negatively) but the horizontal velocity is constant.

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