# Question Video: Calculating the Range of a Projectile Physics • 9th Grade

A projectile has an initial speed of 15 m/s at a launch angle of 28Β° above the horizontal. What is the horizontal displacement of the projectile from its launch position to where it lands if its vertical displacement from its launch position is zero?

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### Video Transcript

A projectile has an initial speed of 15 meters per second at a launch angle of 28 degrees above the horizontal. What is the horizontal displacement of the projectile from its launch position to where it lands if its vertical displacement from its launch position is zero?

Letβs start by drawing a diagram of this scenario. We have a projectile launched at an initial speed that we will call π. The projectile is launched at an angle above the horizontal that we will call π. During projectile motion, the only force acting on the projectile is gravity. So there is a force acting downwards on the projectile which has a magnitude of the mass of the projectile, which we will call π, multiplied by the acceleration due to gravity, which is π.

Because of the downwards force acting on the projectile, its path will be curved. And the question tells us that when the projectile lands, its vertical displacement from its launch position is zero. This means that the final position of the projectile is at the same height as its initial position.

The question asks us to work out what the horizontal displacement of the projectile from its launch position is when it lands. This is also known as the projectileβs range, which we will call capital π. Because there is no force acting on the projectile in the horizontal direction, at any point during the motion, its horizontal displacement that we will call π π₯ is equal to its horizontal velocity, which we will call π π₯, multiplied by time, which we will call π‘.

When the projectile lands, it has a horizontal displacement of capital π. And this occurs at a time that we will call capital π. So we can write π is equal to π π₯ multiplied by capital π. And capital π is also known as the time of flight of the projectile. So to calculate the horizontal range of the projectile capital π, we need to calculate its horizontal velocity π π₯ and the time of flight of the projectile capital π. And weβll keep a note of this down here on the left.

Letβs start by calculating the horizontal velocity of the projectile. We can do this by looking at a diagram of its initial velocity and the angle that it makes with the horizontal. We can see that it has a horizontal component, which is its horizontal velocity, which we have called π π₯. And it has a vertical component, which is the projectileβs initial vertical velocity, which we will call π π¦. These form a right-angled triangle. So we can write that π π₯ is equal to π multiplied by the cos of π and π π¦ is equal to π multiplied by the sin of π. Both of these will be important to us, so weβll keep a note of them down here on the left

So we have an expression for the horizontal velocity of the projectile. All we need to do now is calculate its time of flight. The time of flight of a projectile that lands at the same height that it was launched from can be calculated using the following formula. Time of flight, which is capital π, is equal to two multiplied by the initial vertical velocity π π¦ divided by the acceleration due to gravity π.

Letβs substitute our expression for π π¦ into this equation. So we can write π equals two multiplied by π sin π divided by π. So we now have an expression for the time of flight of a projectile, which weβll keep a note of down here on the left.

We now have everything we need to calculate the horizontal range of the projectile. So weβll start with our expression for horizontal range. π equals π π₯ multiplied by π. We will then substitute our expressions for π π₯ and π into this equation. So we will write π equals π cos π multiplied by two π sin π divided by π. And as we can see, this part on the left is the horizontal velocity and this part on the right is the time of flight. Letβs rearrange this to make it easier to work with.

First, weβll bring the two outside of the brackets. And weβll do the same for each π. And these become π squared. Then we can take cos π and sin π out of the brackets. And finally, the whole equation is divided by π. And this is our final expression for the horizontal range of the projectile.

All thatβs left for us to do is find our values of π, π, and π. The question tells us that the projectile has an initial speed of 15 meters per second. So we can write π equals 15 meters per second. The question also tells us that the projectile was launched at an angle of 28 degrees above the horizontal. So we can write π equals 28 degrees. And finally, π equals 9.8 meters per second squared.

Now π and π are both in SI units and the cos or sin of an angle has no units, so we donβt need to convert any of these before substituting them in. Continuing, we can write π equals two multiplied by 15 squared multiplied by the cos of 28 degrees multiplied by the sin of 28 degrees divided by 9.8 meters per second squared, which gives π is equal to 19.0 meters.

So the horizontal displacement of the projectile from its launch position to where it lands if its vertical displacement from its launch position is zero is equal to 19.0 meters.