Angled projectile motion

When we have a projectile which is launched at an angle (somewhat upwards, somewhat horizontally), we can treat it as a combination of vertical projectile motion and horizontal projectile motion.

Separating velocity into horizontal and vertical components

Let’s say we know a projectile is launched with an initial velocity of v at an angle of \theta to the horizontal. We can find the horizontal and vertical components of the velocity using trigonometry:

• Horizontal component: v_x = v \cos \theta
• Vertical component: v_y = v \sin \theta

Horizontal component

The horizontal component of the motion will be the same as in horizontal projectile motion, so we can use the same equations and values as in that chapter to solve problems involving the horizontal component of the motion:

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 values can be calculated in a similar way to in the vertical projectile motion chapter, but the object will likely fall further than it originally rose, because it was launched off a cliff, for example.

(It may even fall less far than it rose, if it was launched from a hole in the ground, onto the surface, for example.)

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 u_v (the vertical component of the initial velocity), and will increase in the negative direction (downwards) as the projectile falls, due to the acceleration of gravity.

Unlike in horizontal projectile motion, the vertical velocity won’t start at 0ms^{-1}, because the projectile is launched somewhat upwards at an angle, so it has an initial vertical velocity.

That means that the vertical velocity will start positive, and will decrease to 0ms^{-1} at the maximum height, before increasing in the negative direction (downwards) as the projectile falls, due to the acceleration of gravity.

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

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

Displacement

We can use \text{displacement} = \text{initial velocity} \times \text{time} + \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=u_v (the vertical component of the initial velocity, since the projectile is launched at an angle, so it has an initial vertical velocity) instead of u=0ms^{-1} (since the projectile is launched horizontally, so no initial vertical velocity) as in horizontal projectile motion.

Time

Time is the same as in horizontal projectile motion. Just calculate it using our known values for:

• u_v (the vertical component of the initial velocity)
• a (the acceleration due to gravity, which is -9.81ms^{-2})
• v (the vertical velocity at the end of the motion, which will be negative, because it’s falling downwards)

t = \frac{v - u_v}{a}

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