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

Lesson Explainer: Vector Methods with Projectiles Mathematics

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=𝑠𝑖+𝑠𝑗,𝑣=𝑣𝑖+𝑣𝑗,𝑎=𝑎𝑖+𝑎𝑗, 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 𝑢 and constant acceleration vector 𝑎, then its displacement vector 𝑠 at time 𝑡 is given by 𝑠=𝑢𝑡+12𝑎𝑡=𝑢𝑖+𝑢𝑗𝑡+12𝑎𝑖+𝑎𝑗𝑡 or by 𝑠=12𝑢+𝑣𝑡=12(𝑢+𝑣)𝑖+𝑢+𝑣𝑗𝑡.

The particle’s velocity 𝑣 at time 𝑡 is given by 𝑣=𝑢+𝑎𝑡=𝑢𝑖+𝑢𝑗+𝑎𝑖+𝑎𝑗𝑡.

In the special case of a particle moving freely under gravity, that is, with a horizontal acceleration of zero (𝑎=0), we have the additional equation 𝑣=𝑢+2𝑎𝑠.

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 𝑢=6𝑖+30𝑗ms and moves freely under gravity. If 𝑔=9.8ms, 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=𝑠𝑖+𝑠𝑗, 𝑢=𝑢𝑖+𝑢𝑗, and 𝑎=𝑎𝑖+𝑎𝑗, 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 9.8 m⋅s−2 vertically. Therefore, we have 𝑠=𝑢𝑡+12𝑎𝑡=6𝑖+30𝑗×512×9.8𝑗×5=30𝑖+150𝑗122.5𝑗=30𝑖+27.5𝑗.

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

As for the velocity, we have 𝑣=𝑢+𝑎𝑡=6𝑖+30𝑗9.8𝑗×5=6𝑖19𝑗.

Recalling that the speed is the magnitude of the velocity, the speed is 𝑣=6+19=397, which is 19.9 m⋅s−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 𝑢𝑖+𝑢𝑗 m⋅s−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 𝑢 and the projectile’s path’s greatest height . Take 𝑔=9.8ms.

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=𝑠𝑖+𝑠𝑗, 𝑢=𝑢𝑖+𝑢𝑗, and 𝑎=𝑎𝑖+𝑎𝑗, respectively, before applying the laws of motion in their vector forms.

Consider the vector law of motion governing the particle’s displacement, 𝑠=𝑢𝑡+12𝑎𝑡. Substituting in 𝑢=𝑢=𝑢, 𝑎=0, and 𝑎=9.8, we get 𝑠𝑖+𝑠𝑗=𝑢𝑖+𝑢𝑗𝑡+12𝑎𝑖+𝑎𝑗𝑡=𝑢𝑖+𝑢𝑗×𝑡120𝑖+9.8𝑗×𝑡.

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 𝑢 from the two equations 𝑢𝑡=360 and 𝑢𝑡4.9𝑡=0.

Substituting 𝑢𝑡=360 from the first equation into the second, we have 3604.9𝑡=0𝑡=3604.9𝑡=3604.9.

Taking this value of 𝑡 and substituting back into 𝑢𝑡=360, we get 𝑢3604.9=360𝑢=360÷3604.9=42.

Therefore, the initial velocity vector of the particle is 𝑢𝑖+𝑢𝑗=42𝑖+42𝑗msms.

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=𝑢+𝑎𝑡𝑣𝑖+𝑣𝑗=𝑢𝑖+𝑢𝑗+𝑎𝑖+𝑎𝑗𝑡=42𝑖+42𝑗9.8𝑗×𝑡.

We want to calculate the time when the coefficient of 𝑗 is zero, so we have 429.8𝑡=0𝑡=429.8.

Keeping this value in its exact form, we can substitute this time, 𝑡, back into the vertical component of the particle’s displacement to find its greatest height, : =𝑢𝑡12𝑔𝑡=42429.84.9429.8=18090=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 10𝑖+10𝑗, 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 m⋅s−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=𝑠𝑖+𝑠𝑗, 𝑢=𝑢𝑖+𝑢𝑗, and 𝑎=𝑎𝑖+𝑎𝑗, 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=10𝑖+10𝑗. Let us write down the various components of the particle’s motion.

In the horizontal direction, we have 𝑢𝑣=𝑢,𝑠=10,𝑎=0,(tobecalculated)(becausehorizontalaccelerationiszero) and in the vertical direction, we have 𝑢𝑣=0,𝑠=10,𝑎=9.8.(tobecalculated)(becausethemotionishorizontalatthistime)

Let us first calculate the initial vertical velocity using the law of motion 𝑣=𝑢+2𝑎𝑠 for a particle moving freely under gravity: 𝑣=𝑢+2𝑎𝑠0=𝑢+2×(9.8)×10𝑢=196𝑢=14.

We can now calculate the time 𝑡 at which the particle passes through the point (10,10) using 𝑣=𝑢+𝑎𝑡0=149.8𝑡𝑡=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=12𝑢+𝑣𝑡 to find the initial horizontal velocity 𝑢: 𝑠=12(𝑢+𝑣)𝑡10=12(𝑢+𝑣)149.8, and since 𝑢=𝑣, we have 10=𝑢149.8𝑢=10÷149.8=7.

Since we have shown in our working that 𝑢=7 and 𝑢=14, we conclude that the velocity with which the particle left 𝑂 was 𝑢=7𝑖+14𝑗ms.

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 17𝑖+9𝑗 m relative to the origin 𝑂. After 2 seconds, it was at a point 32𝑖+50𝑗 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 m⋅s−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=𝑠𝑖+𝑠𝑗, 𝑢=𝑢𝑖+𝑢𝑗, and 𝑎=𝑎𝑖+𝑎𝑗, 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 32𝑖+50𝑗 but rather its displacement from 𝑃, (3217)𝑖+(509)𝑗=15𝑖+41𝑗. We can work out the particle’s initial velocity using the law of motion 𝑠=𝑢𝑡+12𝑎𝑡.

In the horizontal direction at 2 s, we have a displacement of 15 m and zero acceleration and so 𝑠=𝑢𝑡15=2𝑢𝑢=7.5.

In the vertical direction, we have a displacement of 41 m and an acceleration of 9.8, so 𝑠=𝑢𝑡+12𝑎𝑡41=2𝑢4.9×42𝑢=41+19.6𝑢=30.3.

Therefore, the initial velocity of the particle is 𝑢=7.5𝑖+30.3𝑗ms.

We can now use this velocity vector to find the displacement of the particle from 𝑃 after 7 s using 𝑠=𝑢𝑡+12𝑎𝑡 again. Horizontally, we have 𝑠=𝑢𝑡=7.5×7=52.5 and vertically, we have 𝑠=𝑢𝑡+12𝑎𝑡=30.3×74.9×7=28.

Therefore, the displacement of the particle from 𝑃 after 7 s is 𝑠=52.5𝑖28𝑗, giving us a distance from 𝑃 of 𝑠=52.5+28=59.5m.

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 𝑢=12𝑖+15𝑗 m⋅s−1, as shown in the diagram below. After 𝑇 seconds, the projectile’s velocity is 12𝑖29.1𝑗 m⋅s−1. Taking 𝑔=9.8ms, 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=𝑠𝑖+𝑠𝑗, 𝑢=𝑢𝑖+𝑢𝑗, and 𝑎=𝑎𝑖+𝑎𝑗, 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=𝑢+𝑎𝑡. We have 𝑣=𝑢+𝑎𝑇29.1=159.8𝑇𝑇=44.1÷9.8=4.5.

So, the time at which the projectile’s velocity reaches 12𝑖29.1𝑗 m⋅s−1 is 𝑇=4.5s.

Then, substituting this time into the law of motion 𝑠=12𝑢+𝑣𝑡, we have 𝑠=1212𝑖+15𝑗+12𝑖29.1𝑗×4.5=1224𝑖14.1𝑗×4.5=54𝑖31.725𝑗.

This gives us the particle’s position vector relative to 𝑂 after 4.5 s as 54𝑖31.725𝑗 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=𝑠𝑖+𝑠𝑗, 𝑢=𝑢𝑖+𝑢𝑗, and 𝑎=𝑎𝑖+𝑎𝑗 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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