# Relationship between height and distance in projectile motion

### Projectile Relationships - Delsea 1st Year Physics In this activity you will be investigating the relationships between some of the different variables involved in projectile motion. a data table the height from which you fired the projectile, the horizontal distance it travelled, and the time of flight. Suppose a projectile is thrown from the ground level, then the range is the distance between the launch point and the landing point, where the projectile ( We ignored the height of the launch point.) projectile must satisfy this relationship. What factors influence the trajectory (flight path) of a projectile? projection angle - the Maximum height (m). Range (distance) (m) path) of a projectile? relative projection height - the difference between projection height and landing height.

We're dealing with just the vertical dimension here. So we're just dealing with the vertical. And remember, it's negative because our displacement is going to be downwards. So our vertical velocity-- we already figured that out. It is 90 times the sine of 53 degrees. Actually, let me do it in that same color. The first time we do a new type of problem, it's good to know-- so 90 times the sine of 53 degrees, times our change in time, is equal to the acceleration of-- due to the force of gravity for an object in free fall is going to be negative 9.

But we're dividing that by 2. So we're going to have minus 4. Times our change in time squared. So how do we solve something like this? You can't just factor out a t and solve it. But you might recognize this is a quadratic equation right over here. And the way you solve quadratic equations is you get everything onto one side of this equation. And then you either factor it out.

But more likely in this situation, we will use the quadratic formula, which we've proved in other videos and hopefully given you the intuition for it-- to actually solve for the times where your elevation, where your displacement in the vertical direction is negative 16 meters.

And you'll get two solutions here. And one of the solutions will be a negative change in time. So it'll be like at sometime in the past, you were also at negative 16 meters. That's nonsensical for this problem. So we'll want to take the positive value here. So let's put all of this on one side of the equation.

So let's add 16 to both sides. On the left-hand side, you just get a 0. I'll write the highest degree term first. And then we have plus 90 sine of 53 degrees times delta t, and then plus I'm going to do that in yellow.

All of this is equal to 0. And this, once again, is just a quadratic equation. We can find its roots. And the roots will be in terms of delta t. We can solve for delta t using the quadratic formula. So we get delta t-- and if this is very unfamiliar to you, review the videos on Khan Academy in the algebra playlist on the quadratic formula. And if you don't know where it came from, we also prove it for you. So it's equal to negative B.

B is this term right here, the coefficient on the delta t. So it's going to be negative 90 sine of 53 degrees. I'll write the quadratic formula up here for those of you who don't fully remember it. So if I'm trying to solve Ax squared plus Bx plus C is equal to 0, the roots over here are going to be negative B plus or minus the square root of B squared minus 4AC, all of that over 2 times A.

These are going to be the x values that satisfy this equation up here. So that's all I'm doing over here.

## Homework Help: Relationship between Launch height and range of a projectile

This is the B value. Negative B plus or minus-- and it's going to turn out that we only care about the plus one, because that's going to give us the positive value.

But I'll just write it out here. Plus or minus the square root of B squared. So it's this quantity squared. So it is 90 sine of 53 degrees squared, minus we're going to run [? So minus 4 times A, which is negative 4. Well, let me just write negative 4. C over here is 16, times Let me put this radical all the way over here.

All of that over 2A. So A is negative 4. So now we can get the calculator out to figure out our change in time. And I'm just going to focus on the positive version of it.

## Projectile Motion

I'll leave it up to you to find the negative version and see that it'll give you a negative value for change in time.

And that's nonsensical, so we only care about the positive change in time where we get to a displacement of negative 16 meters. Let's get the calculator out. So we get-- want to do this carefully. We have negative 90 sine of 53 degrees plus. I'm doing the plus version here because that'll give us the positive value-- plus the square root of. And I'll do this in parentheses.

That's that part right there. These two negatives cancel out. So I could say this is plus 4 times positive 4. So plus 4 times 4. And then that closes off our entire radical. And so this will give me the numerator up here. That gives me the numerator. And then I want to divide that by-- did I do the negative 90? Oh, and I just realized that I made a mistake. I said that the positive version would give you the positive time.

But now we realize that's wrong. Because when I took the positive version, when I put a plus up here, I get a positive 2. But then we divide it by negative 9. We're going to get a negative value.

So that's not going to be the time that we care about. We care about the time where this is a negative value. So let me re-enter that. So let me do the negative value. So let me move back a little bit.

Physics - Mechanics: Motion In Two-Dimensions: Projectile Motion (3 of 21) Projectile Upward Angle

And then let me replace this with a minus sign. So I'm going to look at the negative value, because I want the positive time. And so now my numerator here is a negative value. And so this is actually what we care about. Does the speed of the ball hit zero when the ball reach its highest height?

If not, why not?

### Projectile motion - Wikipedia

Does the speed of the ball when it reaches the ground equal the speed of the ball at launch? If not, why not?. These variables will not be altered for the rest of this part of the experiment. For each speed record in a data table the speed you fired the projectile with, the horizontal distance it travelled, and the time of flight. In Logger Pro, create graphs of time of flight vs.

Transfer both graphs to your lab book and make sure they have all the things a good graph should contain. Give a concluding sentence or two for this part of the lab. Fire the projectile at 8 different heights including a height of 0 m.

### Horizontal and Vertical Displacement

For each height record in a data table the height from which you fired the projectile, the horizontal distance it travelled, and the time of flight. For each angle record in a data table the angle from which you fired the projectile, the horizontal distance it travelled, and the time of flight.

Transfer both graphs to your lab book and make sure they have all the things a good graph should contain except for equations. You should not try to curve fit these graphs.

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