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## Class 11 Physics (India)

### Unit 8: Lesson 7

Projectiles launched from/to a height

# Total displacement for projectile

Reconstructing the total displacement vector for a projectile. Created by Sal Khan.

## Want to join the conversation?

• How would you know wheather or not the projectile hit the wall? • does sal have video on introduction to vectors with all the laws, parrelleelgram law, product of vector,etc? • So what if the base distance had been something less than 2 meters, something like 1.5 meters, in which case the projectile would crash into the side of the wall? How would you figure out vertical and horizontal displacement then? And even more challenging, what about the velocity and acceleration at the point of impact? • At , why do we know that displacement is 10? How can we assume that the horizontal component is large enough so that the ball will land on the ledge? It could just as easily land on the ground, making the displacement 0. • Sal is referring to the total vertical displacement of the projectile at this point. As to your second question, you are correct. You cannot assume this without working through the problem. In fact, given the distance and height parameters, we could as easily chosen an initial velocity that would have caused the projectile to miss the ledge. I think the initial velocity of the problem in the video was probably chosen with some knowledge (or educated suspicion) that it would make it past the ledge.
• realistically, wouldnt there be two times that the cannon ball passes 10m ? what if you wanted to find out the first time that the ball passes 10 m and then comes back down again to 10m? • In addition to understand this kind of movement, if you solve both values of the quadratic equation, you will get two "positive(+)" answers. As the teacher said, time can't and never be "negative(-)", we don't have any (-) magnitude for "time", we have 2 (+) values and the one teacher's solved is the second one which is the exact time when it reaches the landing. However, the other value of time is when it crosses 10m altitude from the ground.
Hope it was helpful, GL fellow students :) • How do we find out the parabolic distance covered by the object? • You could use the trajectory formula. If you graph the result you'll see the entire distance traveled by the object throughout the range, or total horizontal displacement. Plugging in any values from the initial horizontal position to the final horizontal position(delta x, a.k.a. R, or range) will give you the change in height at that, and hence if you put in the total change in the horizontal position(the range) you get a graph of the vertical displacement vs the horizontal displacement. Which is basically like a map of the path the projectile takes in real life! And the path a projectile takes in space(this includes the y AND x directions, hence why you get a graph that is y vs x) is called its trajectory, and if you are wondering what its formula is, just search it up on google, it's pretty long. IF you want to know the magnitude of the total displacement vector, r, which is what it sounds like you are asking, that's even easier. For a projectile where the change in the vertical displacement is 0, thanks to no change in elevation from the starting vertical position to the final one, then the displacement vector r is equal to the magnitude of delta x, the change in horizontal displacement. Its direction is the same as that as well. For a projectile where there is a change in elevation, just apply Pythagorean theorem of the x and y displacement components and add the squares of both of them then take the square root to find the magnitude of the total displacement vector r. As for the direction, divide the y component of the displacement, delta y, by the x component of displacement, delta x, then find the inverse tangent to get the angle, which is the same as the direction, of the displacement vector r. :) so no need for calculus for that stuff.
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• How would you find initial velocity of an object if only given 1: the height of the building it is thrown from and 2: the total displacement of the object? I'm really lost on this one....help! ex A stone thrown horizontally from the top of a 34.0-m tower hits the ground at a point 25.0 m from the base of the tower. Argg! • Divide the movement into a horizontal and a vertical movement. The vertical movement is the stone dropping with gravitational acceleration for 34 m (heigth of tower). From this you can calculate the time. During the same time, the stone also moves horizontally at its inital velocity. Neglecting air friction, this velocity remains constant. So take the formula for something moving at constant velocity. You got the distance already and you get the time from solving the vertical movement equation. The result should be v0 = 13.4 m/s if I am not mistaken.  