Lesson Explainer: Vector Methods with Projectiles | Nagwa Lesson Explainer: Vector Methods with Projectiles | Nagwa

Lesson Explainer: Vector Methods with Projectiles Mathematics

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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 ⃑𝑠=𝑠⃑𝑖+𝑠⃑𝑗,⃑𝑣=𝑣⃑𝑖+𝑣⃑𝑗,βƒ‘π‘Ž=π‘Žβƒ‘π‘–+π‘Žβƒ‘π‘—,ο—ο˜ο—ο˜ο—ο˜ where ⃑𝑖 and ⃑𝑗 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 ⃑𝑒 and constant acceleration vector βƒ‘π‘Ž, then its displacement vector ⃑𝑠 at time 𝑑 is given by ⃑𝑠=⃑𝑒𝑑+12βƒ‘π‘Žπ‘‘=𝑒⃑𝑖+𝑒⃑𝑗𝑑+12ο€Ίπ‘Žβƒ‘π‘–+π‘Žβƒ‘π‘—ο†π‘‘οŠ¨ο—ο˜ο—ο˜οŠ¨ or by ⃑𝑠=12⃑𝑒+⃑𝑣𝑑=12ο€Ί(𝑒+𝑣)⃑𝑖+𝑒+𝑣⃑𝑗𝑑.ο—ο—ο˜ο˜

The particle’s velocity 𝑣 at time 𝑑 is given by ⃑𝑣=⃑𝑒+βƒ‘π‘Žπ‘‘=𝑒⃑𝑖+𝑒⃑𝑗+ο€Ίπ‘Žβƒ‘π‘–+π‘Žβƒ‘π‘—ο†π‘‘.ο—ο˜ο—ο˜

In the special case of a particle moving freely under gravity, that is, with a horizontal acceleration of zero (π‘Ž=0), we have the additional equation 𝑣=𝑒+2π‘Žπ‘ .

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 ⃑𝑒=ο€Ί6⃑𝑖+30⃑𝑗⋅ms and moves freely under gravity. If 𝑔=9.8β‹…ms, find the projectile’s 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 ⃑𝑠=𝑠⃑𝑖+π‘ βƒ‘π‘—ο—ο˜, ⃑𝑒=𝑒⃑𝑖+π‘’βƒ‘π‘—ο—ο˜, and βƒ‘π‘Ž=π‘Žβƒ‘π‘–+π‘Žβƒ‘π‘—ο—ο˜, 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 βˆ’9.8 mβ‹…sβˆ’2 vertically. Therefore, we have ⃑𝑠=⃑𝑒𝑑+12βƒ‘π‘Žπ‘‘=ο€Ί6⃑𝑖+30⃑𝑗×5βˆ’12Γ—9.8⃑𝑗×5=30⃑𝑖+150βƒ‘π‘—βˆ’122.5⃑𝑗=30⃑𝑖+27.5⃑𝑗.

So, the position vector of the projectile after 5 seconds is 30⃑𝑖+27.5⃑𝑗.

As for the velocity, we have ⃑𝑣=⃑𝑒+βƒ‘π‘Žπ‘‘=6⃑𝑖+30βƒ‘π‘—βˆ’9.8⃑𝑗×5=6βƒ‘π‘–βˆ’19⃑𝑗.

Recalling that the speed is the magnitude of the velocity, the speed is ‖‖⃑𝑣‖‖=√6+19=√397, which is 19.9 mβ‹…sβˆ’1 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 𝑒⃑𝑖+𝑒⃑𝑗 mβ‹…sβˆ’1 from a fixed point 𝑂 in a horizontal plane landed at a point in the same plane 360 m away. Find the value of 𝑒 and the projectile’s path’s greatest height β„Ž. Take 𝑔=9.8β‹…ms.

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 ⃑𝑠=𝑠⃑𝑖+π‘ βƒ‘π‘—ο—ο˜, ⃑𝑒=𝑒⃑𝑖+π‘’βƒ‘π‘—ο—ο˜, and βƒ‘π‘Ž=π‘Žβƒ‘π‘–+π‘Žβƒ‘π‘—ο—ο˜, respectively, before applying the laws of motion in their vector forms.

Consider the vector law of motion governing the particle’s displacement, ⃑𝑠=⃑𝑒𝑑+12βƒ‘π‘Žπ‘‘οŠ¨. Substituting in 𝑒=𝑒=π‘’ο—ο˜, π‘Ž=0, and π‘Ž=βˆ’9.8, we get 𝑠⃑𝑖+𝑠⃑𝑗=𝑒⃑𝑖+𝑒⃑𝑗𝑑+12ο€Ίπ‘Žβƒ‘π‘–+π‘Žβƒ‘π‘—ο†π‘‘=𝑒⃑𝑖+π‘’βƒ‘π‘—ο†Γ—π‘‘βˆ’12ο€Ί0⃑𝑖+9.8⃑𝑗×𝑑.ο—ο˜ο—ο˜ο—ο˜οŠ¨οŠ¨

When the particle lands, the horizontal component of its displacement (the coefficient of ⃑𝑖) is equal to 360, and the vertical component of its displacement (the coefficient of ⃑𝑗) is equal to 0. We can therefore calculate the value of 𝑒 from the two equations 𝑒𝑑=360 and π‘’π‘‘βˆ’4.9𝑑=0.

Substituting 𝑒𝑑=360 from the first equation into the second, we have 360βˆ’4.9𝑑=0𝑑=3604.9𝑑=ο„ž3604.9.

Taking this value of 𝑑 and substituting back into 𝑒𝑑=360, we get π‘’ο„ž3604.9=360𝑒=360Γ·ο„ž3604.9=42.

Therefore, the initial velocity vector of the particle is 𝑒⃑𝑖+𝑒⃑𝑗⋅=ο€Ί42⃑𝑖+42⃑𝑗⋅msms.

We can now calculate the particle’s 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 ⃑𝑣=⃑𝑒+βƒ‘π‘Žπ‘‘π‘£βƒ‘π‘–+𝑣⃑𝑗=𝑒⃑𝑖+𝑒⃑𝑗+ο€Ίπ‘Žβƒ‘π‘–+π‘Žβƒ‘π‘—ο†π‘‘=42⃑𝑖+42βƒ‘π‘—βˆ’9.8⃑𝑗×𝑑.ο—ο˜ο—ο˜ο—ο˜

We want to calculate the time when the coefficient of ⃑𝑗 is zero, so we have 42βˆ’9.8𝑑=0𝑑=429.8.

Keeping this value in its exact form, we can substitute this time, 𝑑, back into the vertical component of the particle’s displacement to find its greatest height, β„Ž: β„Ž=π‘’π‘‘βˆ’12𝑔𝑑=42ο€Ό429.8οˆβˆ’4.9ο€Ό429.8=180βˆ’90=90.

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 𝑂 passed horizontally through a point with a position vector of ο€Ί10⃑𝑖+10⃑𝑗, where ⃑𝑖 and ⃑𝑗 are horizontal and vertical unit vectors respectively. Determine the velocity with which the particle left 𝑂, considering the acceleration due to gravity to be 9.8 mβ‹…sβˆ’2.

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 ⃑𝑠=𝑠⃑𝑖+π‘ βƒ‘π‘—ο—ο˜, ⃑𝑒=𝑒⃑𝑖+π‘’βƒ‘π‘—ο—ο˜, and βƒ‘π‘Ž=π‘Žβƒ‘π‘–+π‘Žβƒ‘π‘—ο—ο˜, 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 𝑣=0, that is, when it is at its greatest height. We are told that at this time, its position vector is ⃑𝑠=10⃑𝑖+10⃑𝑗. Let us write down the various components of the particle’s motion.

In the horizontal direction, we have 𝑒𝑣=𝑒,𝑠=10,π‘Ž=0,(tobecalculated)(becausehorizontalaccelerationiszero) and in the vertical direction, we have 𝑒𝑣=0,𝑠=10,π‘Ž=βˆ’9.8.(tobecalculated)(becausethemotionishorizontalatthistime)

Let us first calculate the initial vertical velocity using the law of motion 𝑣=𝑒+2π‘Žπ‘ οŠ¨ο˜οŠ¨ο˜ο˜ο˜ for a particle moving freely under gravity: 𝑣=𝑒+2π‘Žπ‘ 0=𝑒+2Γ—(βˆ’9.8)Γ—10𝑒=196𝑒=14.

We can now calculate the time 𝑑 at which the particle passes through the point (10,10) using 𝑣=𝑒+π‘Žπ‘‘0=14βˆ’9.8𝑑𝑑=149.8.

Keeping this value for the time in its exact form, we can substitute it into the horizontal component of the law of motion ⃑𝑠=12⃑𝑒+⃑𝑣𝑑 to find the initial horizontal velocity 𝑒: 𝑠=12(𝑒+𝑣)𝑑10=12(𝑒+𝑣)149.8, and since 𝑒=𝑣, we have 10=𝑒149.8𝑒=10Γ·149.8=7.

Since we have shown in our working that 𝑒=7 and 𝑒=14, we conclude that the velocity with which the particle left 𝑂 was ⃑𝑒=ο€Ί7⃑𝑖+14⃑𝑗⋅ms.

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 𝑃 with a position vector ο€Ί17⃑𝑖+9⃑𝑗 m relative to the origin 𝑂. After 2 seconds, it was at a point ο€Ί32⃑𝑖+50⃑𝑗 m relative to 𝑂, where ⃑𝑖 and ⃑𝑗 are the horizontal and vertical unit vectors respectively. Calculate the distance of the particle from 𝑃 after a further 5 seconds. Take the acceleration due to gravity to be 9.8 mβ‹…sβˆ’2.

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 ⃑𝑠=𝑠⃑𝑖+π‘ βƒ‘π‘—ο—ο˜, ⃑𝑒=𝑒⃑𝑖+π‘’βƒ‘π‘—ο—ο˜, and βƒ‘π‘Ž=π‘Žβƒ‘π‘–+π‘Žβƒ‘π‘—ο—ο˜, respectively, before applying the laws of motion in their vector forms.

Let us sketch the situation.

Since we are asked to find the distance from 𝑃, it will be simplest to work relative to 𝑃. That is, we will consider the particle’s displacement after 2 s to be not its absolute position vector 32⃑𝑖+50⃑𝑗 but rather its displacement from 𝑃, (32βˆ’17)⃑𝑖+(50βˆ’9)⃑𝑗=15⃑𝑖+41⃑𝑗. We can work out the particle’s initial velocity using the law of motion ⃑𝑠=⃑𝑒𝑑+12βƒ‘π‘Žπ‘‘οŠ¨.

In the horizontal direction at 2 s, we have a displacement of 15 m and zero acceleration and so 𝑠=𝑒𝑑15=2𝑒𝑒=7.5.

In the vertical direction, we have a displacement of 41 m and an acceleration of βˆ’9.8, so 𝑠=𝑒𝑑+12π‘Žπ‘‘41=2π‘’βˆ’4.9Γ—42𝑒=41+19.6𝑒=30.3.

Therefore, the initial velocity of the particle is ⃑𝑒=ο€Ί7.5⃑𝑖+30.3⃑𝑗⋅ms.

We can now use this velocity vector to find the displacement of the particle from 𝑃 after 7 s using ⃑𝑠=⃑𝑒𝑑+12βƒ‘π‘Žπ‘‘οŠ¨ again. Horizontally, we have 𝑠=𝑒𝑑=7.5Γ—7=52.5 and vertically, we have 𝑠=𝑒𝑑+12π‘Žπ‘‘=30.3Γ—7βˆ’4.9Γ—7=βˆ’28.

Therefore, the displacement of the particle from 𝑃 after 7 s is ⃑𝑠=52.5βƒ‘π‘–βˆ’28⃑𝑗, giving us a distance from 𝑃 of ‖‖⃑𝑠‖‖=√52.5+28=59.5m.

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

Example 5: Finding Distances and Times of a Projected Particle

A projectile is launched from a point 𝑂 on top of a tower with initial velocity ⃑𝑒=ο€Ί12⃑𝑖+15⃑𝑗 mβ‹…sβˆ’1, as shown in the diagram below. After 𝑇 seconds, the projectile’s velocity is ο€Ί12βƒ‘π‘–βˆ’29.1⃑𝑗 mβ‹…sβˆ’1. Taking 𝑔=9.8β‹…ms, find the value of 𝑇 and the position vector of the projectile relative to 𝑂 at time 𝑇.

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 ⃑𝑠=𝑠⃑𝑖+π‘ βƒ‘π‘—ο—ο˜, ⃑𝑒=𝑒⃑𝑖+π‘’βƒ‘π‘—ο—ο˜, and βƒ‘π‘Ž=π‘Žβƒ‘π‘–+π‘Žβƒ‘π‘—ο—ο˜, respectively, before applying the laws of motion in their vector forms.

We can calculate the time 𝑇 using the vertical component of the law of motion ⃑𝑣=⃑𝑒+βƒ‘π‘Žπ‘‘. We have 𝑣=𝑒+π‘Žπ‘‡βˆ’29.1=15βˆ’9.8𝑇𝑇=44.1Γ·9.8=4.5.

So, the time at which the projectile’s velocity reaches ο€Ί12βƒ‘π‘–βˆ’29.1⃑𝑗 mβ‹…sβˆ’1 is 𝑇=4.5s.

Then, substituting this time into the law of motion ⃑𝑠=12⃑𝑒+⃑𝑣𝑑, we have ⃑𝑠=12ο€Ί12⃑𝑖+15⃑𝑗+12βƒ‘π‘–βˆ’29.1⃑𝑗×4.5=12ο€Ί24βƒ‘π‘–βˆ’14.1⃑𝑗×4.5=54βƒ‘π‘–βˆ’31.725⃑𝑗.

This gives us the particle’s position vector relative to 𝑂 after 4.5 s as ο€Ί54βƒ‘π‘–βˆ’31.725⃑𝑗 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 ⃑𝑠=𝑠⃑𝑖+π‘ βƒ‘π‘—ο—ο˜, ⃑𝑒=𝑒⃑𝑖+π‘’βƒ‘π‘—ο—ο˜, and βƒ‘π‘Ž=π‘Žβƒ‘π‘–+π‘Žβƒ‘π‘—ο—ο˜ 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.

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