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Acute Angle Projections

Neil Trivedi

Teacher

Neil Trivedi

Contents

Acute Angle Projections

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Challenging Question

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Acute Angle Projections

In this note, we’ll explore acute angle projections under the influence of gravity from either the ground or at a height above the ground. Like forces, we do not like to observe speeds at random angles, so we resolve the speed into its horizontal and vertical components.

The diagram below illustrates a general case of projectile motion. A particle is projected from a point , which is metres above level horizontal ground at a fixed origin , with an initial velocity of
ms, and at an angle of elevation .

Projectile motion diagram showing an object launched from height h at angle θ with speed U, following a curved path to the ground at point B.

The horizontal component of the initial velocity is , while the vertical component is . This can be verified using SOHCAHTOA.

Projectile motion diagram showing a launch from height h with speed U at angle θ, resolved into horizontal component Ucosθ and vertical component Usinθ.

Vertical Axis

• For vertical motion, the particle moves under the influence of gravity. We take the positive direction as the direction of projection, which is usually upwards. Therefore, acceleration due to gravity is taken as negative: ms.

• We model the particle’s vertical motion using the SUVAT equations.

Horizontal Axis

• For horizontal motion, we ignore any external forces such as air resistance and rotational forces. Using Newton’s second law, , and noting that there are no external horizontal forces, we have (the mass cannot be ). This means that the horizontal velocity remains constant because the horizontal acceleration is .

• We model the particle’s horizontal motion using , where is the distance, is the constant speed, and is the time.

Example 1:

A particle is projected from a point , on level horizontal ground, with speed ms and at an angle of elevation of . The particle moves freely under gravity.

Find the length of time for which the particle is m or more above .

Projectile motion diagram showing an object launched from point O at 60m s−1 and angle 60^∘, following a curved trajectory before descending.

Step 1: Annotate the diagram by resolving and adding extra information from the question.

In the diagram, we resolve the velocity into its vertical and horizontal components.

Projectile motion diagram showing an object launched at 60m s−1 and 65^∘, with horizontal and vertical velocity components and two points where the projectile reaches a height of 100m.

Step 2: Label the vertical and horizontal axes separately. The vertical motion is modelled by SUVAT, and the horizontal motion is modelled by Distance Speed Time.

Let be the times at which the particle is m above the ground.

VerticalHorizontal





Note: Time is what will always connect
the vertical and horizontal components
as time is dimensionless.

Step 3: Use SUVAT to find the two times at which the particle is m above and the length of time the particle is m or more above is the difference between these two times.

Only the vertical components give us enough information to solve this problem so we can ignore the horizontal plane.

Use the quadratic formula to solve for .

Therefore, the length of time the particle is m or more above is seconds to significant figures.

No answer provided.

Example 2:

A particle is projected from a point , which is m above level horizontal ground, with speed ms at an angle of elevation , where . The particle’s path is shown in the diagram and moves freely under gravity. The particle passes through a point , which is m above the ground, and hits the ground at a point .

Projectile motion diagram showing an object launched from a height of 35m at 40 m/s and angle θ, passing point B at a height of 25m before landing at point C.

a) Find the time of flight from to .

Step 1: Resolve the vertical and horizontal components of the initial velocity.

The horizontal component is and the vertical component is (the diagram below illustrates this). Then, we find and using SOHCAHTOA. The triangle helps us visualise this too.

Projectile motion diagram showing a ball launched from a height of 35m at 40 m/s and angle θ, with velocity resolved into horizontal and vertical components as it passes point B at 25m before landing at C.

Given we can use SOHCAHTOA and the triangle to find that and . Therefore:

Right-angled triangle with side lengths 3, 4, and 5, showing angle θ between the base and hypotenuse.

Projectile motion diagram showing a ball launched from a height of 35m at 40 m/s, with horizontal and vertical velocity components of 24 m/s and 32 m/s, passing point B at 25m before landing at C.

Step 2: Label the vertical and horizontal axes separately. The vertical motion is modelled by SUVAT, and the horizontal motion is modelled by Distance Speed Time.

VerticalHorizontal





Note: Since we are taking upwards to be positive, the particle is m below our point of projection, which is why we are setting the displacement here to be negative. Think of it as your point of projection is the origin of an axes and where it hits the ground has a negative
value.

Step 3: Use SUVAT to work out the value of .

Solve for using the quadratic formula.

cannot be negative so the time of flight from to is approximately seconds to two significant figures.


b) Find the distance .

Single Step: Find the horizontal distance using .

m

Use the rounded value from part a) for time, not the exact
value calculated when computing the quadratic formula.


c) Find the speed of the particle at .

The following diagram illustrates the particle’s motion at .

Projectile motion diagram showing a ball launched from a height of 35m at 40 m/s with velocity components 24 m/s horizontally and 32 m/s vertically, and showing the horizontal and vertical velocity components again at point B during descent.

Step 1: The horizontal component of the velocity is constant at ms, so we only need to find the vertical component, denoted by , at this point. Use SUVAT.

At the particle is m above horizontal ground which is m below the point of projection so its displacement is m.

As is pointing downwards, so the velocity is ms. However, we’re looking for the speed, so the direction doesn’t matter, only the magnitude.

Step 2: To find the speed at , we use Pythagoras’ Theorem with the horizontal and vertical components of the velocity.

ms


d) Find the angle that the velocity of the particle at makes with the horizontal.

Single Step: Use SOHCAHTOA to find the angle.

The following diagram illustrates the angle that the velocity makes with the horizontal.

Velocity vector diagram at point B showing horizontal velocity 24 and downward vertical velocity √1220, with angle θ between the resultant velocity and the horizontal.

Therefore, the velocity of the particle at is in the direction of approximately below the horizontal.

No answer provided.

Challenging Question

Practice Questions

Horizontal Projections Projectile Motion and Vectors