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Lesson Explainer: Projectile Motion at Any Angle Mathematics

In this explainer, we will learn how to find the horizontal and vertical components of the velocity of a projectile and analyze and solve problems associated with projectile motion at any angle.

Suppose a particle is projected from a flat horizontal plane, at some angle π›Όβˆ˜ above the horizontal, with an initial velocity of π‘ˆ mβ‹…sβˆ’1.

To analyze this motion, it can be very useful to split the particle’s velocity up into its horizontal and vertical components.

The horizontal and vertical components of the velocity are given by π‘ˆ=π‘ˆ(𝛼),π‘ˆ=π‘ˆ(𝛼).ο—ο˜cossin

We write ⃑𝑖 for the unit length vector in the horizontal direction and ⃑𝑗 for the perpendicular unit length vector in the vertical direction. Then, we can organize this information into a single velocity vector expressed in terms of the unit vectors ⃑𝑖 and ⃑𝑗: βƒ‘π‘ˆ=π‘ˆβƒ‘π‘–+π‘ˆβƒ‘π‘—=π‘ˆ(𝛼)⃑𝑖+π‘ˆ(𝛼)⃑𝑗.ο—ο˜cossin

On the other hand, if we are given a projectile’s initial velocity vector, we can recover its initial speed and angle of projection.

If a projectile has an initial velocity vector π‘ˆβƒ‘π‘–+π‘ˆβƒ‘π‘—ο—ο˜, then applying the Pythagorean theorem gives us the initial speed as π‘ˆ=ο„π‘ˆ+π‘ˆ.οŠ¨ο—οŠ¨ο˜

Looking at the diagram of horizontal and vertical components of the initial velocity, we can see that tan(𝛼)=π‘ˆπ‘ˆο˜ο—, and so the angle of projection is 𝛼=ο€Ύπ‘ˆπ‘ˆοŠ.tanοŠ±οŠ§ο˜ο—

Let us practice writing down the initial velocity in vector form.

Example 1: Finding the Vector Form of the Initial Velocity of a Vector

A particle is projected from a point on a horizontal plane with an initial velocity of 39 mβ‹…sβˆ’1 at an angle of πœƒ above the horizontal, where tan(πœƒ)=512. Express the initial velocity as a vector in terms of ⃑𝑖 and ⃑𝑗, the horizontal and vertical unit vectors in the vertical plane.

Answer

The horizontal and vertical components of the initial velocity vector βƒ‘π‘ˆ=π‘ˆβƒ‘π‘–+π‘ˆβƒ‘π‘—ο—ο˜ are given by π‘ˆ=π‘ˆ(πœƒ),π‘ˆ=π‘ˆ(πœƒ),ο—ο˜cossin where πœƒ is the angle of projection. In this question, we are not given the value of πœƒ, but rather its tangent. However, this is enough information to write down sin(πœƒ) and cos(πœƒ), which are what we need here. Given that tan(πœƒ)=512, we can sketch the following triangle.

The Pythagorean theorem gives us the length of the hypotenuse as β„Ž=√12+5=√169=13.

Therefore, sin(πœƒ)=513 and cos(πœƒ)=1213. Furthermore, we are told that the initial velocity of the particle is 39 mβ‹…sβˆ’1, so our initial velocity vector is βƒ‘π‘ˆ=π‘ˆ(πœƒ)⃑𝑖+π‘ˆ(πœƒ)⃑𝑗=39Γ—1213⃑𝑖+39Γ—513⃑𝑗=36⃑𝑖+15⃑𝑗⋅.cossinms

Having decomposed an initial velocity into horizontal and vertical components, let us now see an example of the opposite procedure: recovering the initial velocity and angle of projection from a velocity vector.

Example 2: Finding the Initial Speed and Angle of Projection of a Projectile given Its Initial Velocity Vector

  1. A particle is projected with velocity βƒ‘π‘ˆ=4⃑𝑖+7⃑𝑗⋅ms, where ⃑𝑖 and ⃑𝑗 are horizontal and vertical unit vectors. Find the initial speed of the particle approximated to one decimal place.
  2. Find the angle of projection of the particle approximated to one decimal place.

Answer

Part 1

If a projectile has an initial velocity vector π‘ˆβƒ‘π‘–+π‘ˆβƒ‘π‘—ο—ο˜, then the initial speed is given by π‘ˆ=ο„π‘ˆ+π‘ˆοŠ¨ο—οŠ¨ο˜ and the angle of projection is 𝛼=ο€Ύπ‘ˆπ‘ˆοŠ.tanοŠ±οŠ§ο˜ο—

It is almost always a good idea to draw a sketch. So, let us do that as follows.

From the sketch, we can see that the initial velocity π‘ˆ is given by π‘ˆ=ο„π‘ˆ+π‘ˆ=√4+7=√65=8.1β‹…,οŠ¨ο—οŠ¨ο˜οŠ¨οŠ¨οŠ±οŠ§ms approximated to one decimal place.

Part 2

Furthermore, the angle of projection of the particle is 𝛼=ο€Ύπ‘ˆπ‘ˆοŠ=ο€Ό74=60.3,tantanοŠ±οŠ§ο˜ο—οŠ±οŠ§βˆ˜ approximated to one decimal place.

Now that we know how to decompose a projectile’s velocity into horizontal and vertical components, we can use this technique to solve problems about projectile motion. To do this, we will need to recall the equations of motion. In particular, to solve problems involving the displacement 𝑠 of a projectile at a time 𝑑, given that it has initial velocity 𝑒 and acceleration π‘Ž, we need the equation of motion 𝑠=𝑒𝑑+12π‘Žπ‘‘.

Example 3: Finding the Components of the Initial Velocity of a Projectile

A flare was fired from a flare gun. After 2 seconds, the flare’s horizontal and vertical distances from the flare gun were 22 m and 31 m respectively. Find the horizontal ⃑𝑣 and vertical βƒ‘π‘£ο˜ components of the flare’s initial velocity. Take 𝑔=9.8/ms.

Answer

We will analyze the horizontal and vertical components of the flare’s motion separately. Let us begin with the horizontal component.

We want to find the flare’s initial horizontal velocity ⃑𝑣. After its launch, there are no forces acting on the flare in the horizontal direction. This means that its horizontal acceleration βƒ‘π‘Žο— is zero. We are told that the particle’s horizontal displacement is 22 m after time 𝑑=2s. We can substitute these values into the equation of motion 𝑠=𝑒𝑑+12π‘Žπ‘‘οŠ¨ to get 22=⃑𝑣×2+12Γ—0Γ—2⃑𝑣=22Γ·2=11.ο—οŠ¨ο—

Therefore, the horizontal component of the flare’s initial velocity is ⃑𝑣=11/ms.

We now analyze the vertical component of the flare’s motion. We are told that the flare’s vertical displacement βƒ‘π‘ ο˜ is 31 m after time 𝑑=2s and that its vertical acceleration βƒ‘π‘Žο— is βˆ’9.8 m/s2 due to gravity. Notice that the negative sign on the flare’s vertical acceleration comes from the fact that this acceleration is in the opposite direction to the flare’s initial vertical velocity βƒ‘π‘£ο˜. We substitute these values into the equation of motion 𝑠=𝑒𝑑+12π‘Žπ‘‘οŠ¨ to get 31=⃑𝑣×2βˆ’12Γ—9.8Γ—22⃑𝑣=31+9.8Γ—2⃑𝑣=(31+19.6)Γ·2=25.3.

Therefore, the vertical component of the flare’s initial velocity is ⃑𝑣=25.3/ms.

We can use these same techniques to solve more complicated problems involving projectile motion.

Example 4: Finding the Horizontal Distance of a Particle to Reach the Ground given Its Position after a Given Time of Projection

A particle was projected from a point 𝑂 on horizontal ground. If, after 8 seconds, its horizontal distance from 𝑂 was 48 m and its height was 98 m, determine how far the particle would be from 𝑂 when it hits the ground. Take 𝑔=9.8/ms.

Answer

We need to tackle this question in three steps. First, we will analyze the vertical component of the particle’s motion and use the values given at time 𝑑=8s to find the vertical component βƒ‘π‘’ο˜ of its initial velocity. Second, we will use the initial vertical velocity that we have calculated to work out the time π‘‘οŠ¨ when the particle hits the ground. Third, we will analyze the horizontal component of the particle’s motion to find its horizontal displacement when it hits the ground.

First, we need to find the vertical component βƒ‘π‘’ο˜ of the particle’s initial velocity. We are told that the particle has a vertical displacement of 98 m after 8 s and that its acceleration due to gravity is 9.8 m/s2 in the negative vertical direction. We substitute these values into the equation of motion 𝑠=𝑒𝑑+12π‘Žπ‘‘οŠ¨ to get 98=⃑𝑒×8βˆ’12Γ—9.8Γ—88⃑𝑒=98+313.6⃑𝑒=(98+313.6)Γ·8=51.45.

We now want to find the time π‘‘οŠ¨ when the particle hits the ground. Let us sketch the situation.

The particle hits the ground when its vertical displacement βƒ‘π‘ ο˜ is zero. We can see that there are two times when the particle has a vertical displacement of zero: its moment of launch 𝑑=0 and the time when it hits the ground π‘‘οŠ¨, which we want to find. We substitute again into the law of motion 𝑠=𝑒𝑑+12π‘Žπ‘‘οŠ¨ using the initial velocity ⃑𝑒=51.45/ms that we have just calculated: 0=51.45Γ—π‘‘βˆ’12Γ—9.8×𝑑𝑑(51.45βˆ’4.9𝑑)=0.

We can see that this quadratic equation has two solutions: one at 𝑑=0 and the second at π‘‘οŠ¨, when 51.45βˆ’4.9𝑑=0. Thus, we have 4.9𝑑=51.45𝑑=10.5.

We can now analyze the horizontal component of the particle’s motion to find its horizontal displacement when it hits the ground. We first need to calculate its initial horizontal velocity ⃑𝑒. We are told that at time 𝑑=8, the particle has a horizontal displacement of 48 m. We also know that its horizontal acceleration is 0. We substitute these values into the equation of motion 𝑠=𝑒𝑑+12π‘Žπ‘‘οŠ¨ to get 48=⃑𝑒×8⃑𝑒=6.

Finally, we substitute this horizontal velocity and the time of landing 𝑑=10.5s back into the equation of motion, observing that the horizontal acceleration remains as 0: 𝑠=𝑒𝑑+12π‘Žπ‘‘=6Γ—10.5=63.

We conclude that the particle would be 63 m from 𝑂 when it hits the ground.

We can also solve problems that do not explicitly involve a projectile’s displacement, but rather its initial velocity 𝑒, final velocity 𝑣, and acceleration π‘Ž. For this, we will need the equation of motion 𝑣=𝑒+π‘Žπ‘‘.

The following is an example of this type.

Example 5: Finding the Time Taken by a Projectile to Reach Its Maximum Height

Suppose a particle is projected at a speed of 21 m/s and an angle of elevation 51∘. How long will it take for the particle to reach its greatest height? Give your answer correct to two decimal places. Take 𝑔=9.8/ms.

Answer

To solve this problem, we use the observation that a projectile reaches its greatest height at the moment when it stops rising and starts fallingβ€”that is, when its vertical velocity is 0. In order to make use of this observation, we first need to calculate the vertical component βƒ‘π‘’ο˜ of the particle’s initial velocity.

The vertical component of the particle’s initial velocity is given by ⃑𝑒=21(51)=16.3200β€¦ο˜βˆ˜sin.

We want to find the time 𝑑 when the particle’s vertical velocity βƒ‘π‘£ο˜ is 0. We substitute our initial velocity ⃑𝑒=16.3200β€¦ο˜ into the equation of motion 𝑣=𝑒+π‘Žπ‘‘, observing that the acceleration due to gravity is negative as it is in the opposite direction to the particle’s initial vertical velocity: 0=16.3200β€¦βˆ’9.8𝑑𝑑=16.3200…9.8=1.665….

We conclude that the particle reaches its greatest height after 1.67 s, approximated to two decimal places.

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

Key Points

  • Given a projectile with initial speed π‘ˆ m/s and angle of elevation π›Όβˆ˜, we can decompose its motion into horizontal and vertical components π‘ˆο— and π‘ˆο˜ by using the formulas π‘ˆ=π‘ˆ(𝛼),π‘ˆ=π‘ˆ(𝛼).ο—ο˜cossin
  • We can organize this information into a single velocity vector expressed in terms of the horizontal and vertical unit vectors ⃑𝑖 and ⃑𝑗: βƒ‘π‘ˆ=π‘ˆβƒ‘π‘–+π‘ˆβƒ‘π‘—=π‘ˆ(𝛼)⃑𝑖+π‘ˆ(𝛼)⃑𝑗.ο—ο˜cossin
  • Given an initial velocity vector βƒ‘π‘ˆ=π‘ˆβƒ‘π‘–+π‘ˆβƒ‘π‘—ο—ο˜, we can recover a projectile’s initial speed π‘ˆ and angle of elevation 𝛼 by using the formulas π‘ˆ=ο„π‘ˆ+π‘ˆοŠ¨ο—οŠ¨ο˜ and 𝛼=ο€Ύπ‘ˆπ‘ˆοŠ.tanοŠ±οŠ§ο˜ο—
  • To solve more complicated problems about projectile motion, we can use the horizontal and vertical decomposition of a projectile’s motion along with the equations of motion 𝑠=𝑒𝑑+12π‘Žπ‘‘οŠ¨ and 𝑣=𝑒+π‘Žπ‘‘.

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