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# Total displacement for projectile

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

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• How would you know wheather or not the projectile hit the wall?
• I calculated the object will reach 10.1m vertical after 2 horizontal meters, just enough to clear the wall. Sorry about the format problems...First I found v vert. using the .38s time at 2m.
v_f=v_i+a_c ∆t and then v_f=28.5m/s +(-9.8m/s^2 )0.38s for v_f=24.8m/s

Next:
v_(final vert)^2=(v_initial )^2+2as: (24.78m/s)^2=(28.5m/s)^2+2(-9.8m/s^2)s
(64.8m^2)/s^2 )=((812.3m^2)/s^2 )+(-19.6m/s^2)s ⃗
-(198m^2)/s^2=(19.6m/s^2)s ⃗
(10.1m)=s ⃗ at 2 horiz. meters of travel at 0.38s. This also checks on a graph of the parabola using Sal's equation in Geo Sketchpad.
• 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.
• You are absolutely right. He's probably worked this problem out ahead of time. But if this were on an exam and you don't know whether it reaches the cliff top or not, you could do a quick check. Check what? One approach would be to see how much time it takes to get the 2 meters to the cliff wall. Vx=30*sin(80) or 5.21m/s. So the 2 m spot is reached in 0.38seconds. OK is it high enough at that point? The y position is (at anytime it's in free fall): Y=Yo+Vyot+(1/2)at^2. Yo is 0 , Vyo is 30*sin(80)=29.5m/s and 'a' is -9.8m/s so when the ball reaches the cliff face at t=0.38sec, Y=11.2-0.7=10.5m so the projectile will be a half meter above the top of the cliff when it crosses the edge of the cliff. (If you are standing there, wear knee pads!)

It's often helpful (if time permits) to check the situation - determine a 'corner case' such as this if it's easier than the ultimate problem. What's another corner case? How about how long it takes to get to the maximum height? That happens when Vo=at. Easy to figure - but the problem is you don't know if the trajectory would run into the cliff before this happens, so this may be an unhelpful 'corner case.'
• does sal have video on introduction to vectors with all the laws, parrelleelgram law, product of vector,etc?
• Geeez, been looking at all the physics video's in search of this, didn't think of checking it at algebra. Thank you oh kind stranger
• 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 the quadratic equation we have 2 possible values, and in this scenario, both add up to positive numbers.

t1 = 0.39s
t2 = 5.69s

This means that it takes 0.39s to get the first 10m while it's still moving upwards, then it would complete the movement and land at the platform at 5.69s.
• 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?
• You need to check if the tangent (relative sin and cos, or x and y direction) is greater then 10/2 (5) - which it is at 5.67
• 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 :)
• Is there ever a situation where you want a negative value on time?
• Yes there is. Suppose you know when the projectile hit the ground, and you want to know when it was thrown. It would not be thrown at a positive time because that would mean it will be thrown in the future. The projectile has already been thrown. The time from now when the projectile is thrown will be a negative value. If you want to know the time of something that happened in the past, you must remember that it will almost always be a negative value.
• 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.
(1 vote)
• 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!