In this explainer, we will learn how to solve projectile motion problems using vectors.

We can analyze the motion of a projectile in the vertical plane by expressing its position, velocity, and acceleration as the vector quantities where and are unit vectors in the horizontal and vertical directions. We will then apply the laws of motion in their vector forms.

### Laws: Vector Forms of Laws of Motion

If a particle is moving with initial velocity vector and constant acceleration vector , then its displacement vector at time is given by or by

The particleβs velocity at time is given by

In the special case of a particle moving freely under gravity, that is, with a horizontal acceleration of zero (), we have the additional equation

We can use these laws to calculate the position and speed of a projectile after a given amount of time has elapsed.

### Example 1: Finding the Position Vector and Speed of a Projected Particle

A projectile is launched from the origin with initial velocity and moves freely under gravity. If , find the projectileβs position vector after 5 seconds and its speed at this time. Give your answers to one decimal place.

### Answer

Recall that we can analyze the motion of a projectile in the vertical plane by expressing its position, velocity, and acceleration as the vector quantities , , and , respectively, before applying the laws of motion in their vector forms.

We calculate the position and velocity vectors of the projectile after
5 seconds
using the vector laws of motion. Starting with the position, note that
the only force acting on the projectile is gravity, so its acceleration is
mβ
s^{β2} vertically. Therefore, we have

So, the position vector of the projectile after 5 seconds is .

As for the velocity, we have

Recalling that the speed is the magnitude of the velocity, the speed is
, which is
19.9 mβ
s^{β1}
to one decimal place.

We can use the fact that a projectile has a vertical displacement of zero when it lands on the horizontal plane from which it was launched to calculate its initial velocity.

We can also use the fact that a projectile has a vertical velocity of zero when it reaches its greatest height to calculate that greatest height. Here is an example that tests these skills.

### Example 2: Finding the Initial Velocity and Greatest Height of a Projected Particle in Vector Form

A particle projected with a velocity mβ
s^{β1}
from a fixed point in a horizontal plane landed at a
point in the same plane
360 m away. Find the value of and the
projectileβs pathβs greatest height . Take
.

### Answer

Recall that we can analyze the motion of a projectile in the vertical plane by expressing its position, velocity, and acceleration as the vector quantities , , and , respectively, before applying the laws of motion in their vector forms.

Consider the vector law of motion governing the particleβs displacement, . Substituting in , , and , we get

When the particle lands, the horizontal component of its displacement (the coefficient of ) is equal to 360, and the vertical component of its displacement (the coefficient of ) is equal to 0. We can therefore calculate the value of from the two equations and

Substituting from the first equation into the second, we have

Taking this value of and substituting back into , we get

Therefore, the initial velocity vector of the particle is .

We can now calculate the particleβs greatest height. Observe that the particle reaches its greatest height when it stops rising and starts falling, that is, when its vertical velocity is zero. Therefore, from the vector laws of motion, we have

We want to calculate the time when the coefficient of is zero, so we have

Keeping this value in its exact form, we can substitute this time, , back into the vertical component of the particleβs displacement to find its greatest height, :

Therefore, the greatest height that the particle reaches is 90 m.

Another type of problem we can solve with these methods is to calculate the initial velocity of a projectile by working back from the fact that it passes through a given point with a given velocity.

### Example 3: Determining the Velocity of a Projected Particle given the Position Vector

A particle projected from the origin passed horizontally through
a point with a position vector of
,
where and
are horizontal and vertical unit
vectors respectively. Determine the velocity with which the particle left
, considering the acceleration due to gravity to be
9.8 mβ
s^{β2}.

### Answer

Recall that we can analyze the motion of a projectile in the vertical plane by expressing its position, velocity, and acceleration as the vector quantities , , and , respectively, before applying the laws of motion in their vector forms.

Let us sketch the situation.

The particle is traveling horizontally when the vertical component of its velocity , that is, when it is at its greatest height. We are told that at this time, its position vector is . Let us write down the various components of the particleβs motion.

In the horizontal direction, we have and in the vertical direction, we have

Let us first calculate the initial vertical velocity using the law of motion for a particle moving freely under gravity:

We can now calculate the time at which the particle passes through the point using

Keeping this value for the time in its exact form, we can substitute it into the horizontal component of the law of motion to find the initial horizontal velocity : and since , we have

Since we have shown in our working that and , we conclude that the velocity with which the particle left was .

We can also use vector methods to calculate the displacement of a particle relative to given points.

### Example 4: Calculating the Distance of a Particle from a Point Based on Time and the Position Vector

A particle was projected from a point with a position vector
m relative to the origin
. After
2 seconds,
it was at a point
m
relative to , where
and are the horizontal and vertical
unit vectors respectively. Calculate the distance of the particle from
after a further
5 seconds. Take the acceleration due to gravity to be
9.8 mβ
s^{β2}.

### Answer

Let us sketch the situation.

Since we are asked to find the distance from , it will be simplest to work relative to . That is, we will consider the particleβs displacement after 2 s to be not its absolute position vector but rather its displacement from , . We can work out the particleβs initial velocity using the law of motion .

In the horizontal direction at 2 s, we have a displacement of 15 m and zero acceleration and so

In the vertical direction, we have a displacement of 41 m and an acceleration of , so

Therefore, the initial velocity of the particle is .

We can now use this velocity vector to find the displacement of the particle from after 7 s using again. Horizontally, we have and vertically, we have

Therefore, the displacement of the particle from after 7 s is , giving us a distance from of .

Our final example shows how we can calculate the time taken for a projectileβs velocity to change and extrapolate its trajectory.

### Example 5: Finding Distances and Times of a Projected Particle

A projectile is launched from a point on top of a tower with
initial velocity
mβ
s^{β1},
as shown in the diagram below. After
seconds,
the projectileβs velocity is
mβ
s^{β1}. Taking ,
find the value of and the position vector of the
projectile relative to at time .

### Answer

We can calculate the time using the vertical component of the law of motion . We have

So, the time at which the projectileβs velocity reaches
mβ
s^{β1} is
.

Then, substituting this time into the law of motion , we have

This gives us the particleβs position vector relative to after 4.5 s as m.

Let us finish by recapping a few important concepts from this explainer.

### Key Points

- We can analyze the motion of a projectile in the vertical plane by expressing its position, velocity and acceleration as the vector quantities , , and respectively.
- We can use the vector forms of the laws of motion to solve problems about projectiles moving freely under gravity.
- A projectile reaches its greatest height when the vertical component of its velocity is zero.
- A particle projected from a horizontal plane has a vertical displacement of zero at its launch and at its landing.