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## Class 11 Physics (India)

### Unit 8: Lesson 6

Projectiles launched at an angle

# Projectile at an angle

Figuring out the horizontal displacement for a projectile launched at an angle. Created by Sal Khan.

## Want to join the conversation?

• Is there any logical explanation for why vertical component of velocity vector is always used to figure out the time and the horizontal component for figuring out the displacement ? •   The only force acting on the projectile is gravity, since we explicitly are ignoring air resistance. Gravity only affects the vertical component of the projectile's travel. So Sal does the calculations to determine the effects of gravity on the vertical component, which will be to slow the vertical climb to zero then accelerate the projectile back to earth. The time for this effect to take place is the length of time of the flight of the projectile.

The distance the projectile travels is determined by the horizontal component of its flight. If you multiply the horizontal speed by time in the air you get the distance traveled.

It's important to realize you can separate the flight of the projectile into its vertical component and horizontal component, solve them separately, and get valid results for the actual flight of the projectile.
• How do you know that the initial vertical velocity and final velocity are equal in magnitude? •  You can derive this yourself: Think about the displacement of a projectile until it is on the ground again. The displacement is the average velocity times change in time. Of course average velocity is the average of the initial velocity and the final velocity. And you know that the total displacement is equal to zero. If you solve this equation for the final velocity, you will see that it is the negative initial velocity, i.e. the same speed, only in the opposite direction.
• why isn't final velocity zero? shouldn't it be 0 as the object comes to a halt? •  The equations that we are using to solve this problem only apply when the projectile is in free fall. This means that the only force acting on it is the force of gravity. So we should only apply them to the motion of the projectile right after it is thrown and right before it hits the ground. Just before it hits the ground, the projectile has some downward speed. Then only after it hits the ground will it have zero velocity, but hitting the ground will introduce another force to this system, and we would need to use more equations to describe its motion. Fortunately, this problem can be solved just with the motion of the projectile before it hits the ground, so we don't need to concern ourselves with anything after that. So we choose the final velocity to be just before it hits the ground.

And what is the final velocity before it hits the ground? Well, the projectile does not lose any energy while from the time right after it is launched to the time just before it lands. We assume this to be true since we are also assuming that there is no air resistance. And since the starting and ending points have the same elevation, we can then assume that the projectile has equal speed at those two points. However, we should easily see that the projectile was at first going up, but then it finishes by going down, thus we have to write the y component of the final velocity with the opposite sign of the y component of the initial velocity. So if the initial velocity is +5, then the final velocity has to be -5.
• When solving for the horizontal displacement why cant we just use
cos30*10=horizontal displacement? •  You can! But the problem is we aren't sure when the ball hits the ground. The projectile question assumes the movement along the x-axis stops when the object touches the ground again (or question will specify what is the displacement upon first hitting the ground)

co30*10 will give us the "speed" along x-axis the ball will move not the total displacement. In this case 8.66m/s. So in 1 second the object would move that far. However its total movement time is dependent on the time the object is in the air. So we would still need to solve for the y-axis for when the displacement for the y-axis is = to 0.
• at , why is the average velocity in the horizontal direction is 5 square roots of 3 metres per second? I know Sal said it is because it doesn't change, but why does it not change? •  Gravity only affects the velocity in the vertical direction, and since we are assuming that there is no air resistance, there is nothing to change the horizontal velocity.
• At approximately why do we say that change in velocity equals acceleration times change in time??.. • The acceleration is what is actually causing the velocity to change, so if you multiply the time by the acceleration, the answer will be how much the acceleration caused the velocity to change (change in velocity)
• What is the relation between the angle of launch and the angle of impact? • If you assume that air resistance is negligible, then the angle of launch and the angle of impact would be the same (If you are landing at the same height).
This is because the horizontal velocity stays the same the whole time, and the vertical velocity at impact is the same as it is at launch (in the opposite direction).
• Question, at , when Sal was getting the displacement equation, shouldnt it have been 5sqrt(3)/2 * time? Because average velocity is final vel + initial vel divided by 2? • He did use the formula you stated. The horizontal velocity is constant. This means that both the final and the initial velocities are equal (equal to 5*sqrt(3)) i.e.

The final velocity = initial velocity = 5*sqrt(3)

So then the average velocity will be = (final vel. + initial vel.)/2
= {5*sqrt(3) + 5*sqrt(3)}/2
= 2*5*sqrt(3)/2
The two '2's will cancel each other out, leaving us with 5*sqrt(3). This is the part that you missed out on while thinking about how Sal did it.

So you'll end up with just 5*sqrt(3)*t for the horizontal displacement of the projectile.

Hope that helped! (If you haven't found the answer already, since this is quite an old question)  