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Class 11 Physics (India)

Unit 8: Lesson 7

Projectiles launched from/to a height

Correction to total final velocity for projectile

Correction to "Total Final Velocity for Projectile" Video. Created by Sal Khan.

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• Sal, why is the final velocity of the projectile not negetive since the final velocity is moving downward from the apex of the projectiles path? Also, you say the angle of the final velocities magnitude is a downward angle of -78.7, again, wouldnt this make the final velocity of the projectile -26.55m/s instead of +26.55m/s as stated in the video?
• He defined '-' in the vertical as downward and '+' in the vertical as upward.
He defined '-' in the horizontal as left and '+' in the horizontal as right.
The final velocity is downward '-' and to the right '+'. Thus, in order to give direction for it, you must instead use an angle and not both '-' and '+'. Thus the || v_f ||, which means magnitude of the final velocity, is 26.55 m/s and it's direction is 78° below the '+' horizontal
• why is the final velocity not equal to 0? There isn't any speed right?
• Velocity in the real world is a 3D velocity with a forward horizontal, a vertical, and a sideways horizontal component.

If you used actual 3D velocity than I don't think you can simplify it to a 2D velocity simply because 3D(such as a cube) does not simplify to 2D unless it is the net of a 3D figure but that is the only way that 3D simplifies to 2D and 3D velocity does not have a net of 2D velocities like how a cube has a net of squares so how could you simplify a 3D velocity problem to a 2D velocity problem?
• Your point is valid, though if you think about it, most motion in our 3d world does more or less happen on a plane. A thrown ball or a bullet or something mostly travels horizontally for 1d, and gravity affects it for a second dimension (unless thrown straight up), but few objects at speed then also move "sidewise". Sure, somewhat, but for many things a 2d assumption is good enough.
• the sum of two vectors has a magnitude of 5 units and direction of 90 . one of the vectors has a magnitude of 2.2 unit and direction of 35 . what are the magnitude and the direction of the second vector? please answer . and thank you .
• the magnitude is 2.8, the direction is 55
(1 vote)
• This is sort of unrelated . . . but isn't Pythagorean's Theorem Pythagorean's Law now, because it has been proven and used correctly over and over for generations?
• No, that's a misunderstanding you have about the meaning of those words theorem and law. A theorem in math is a statement that has been proven on the basis of previously proven statements. Proven. Not hypothesized.
• At in the video, why was the -26.03 written as positive in the opposite side?
• When Sal wrote the opposite side, he was simply referring to the magnitude of that velocity vector. The magnitude is a scalar quantity, and has no sign associated with it. If Sal had written the value as negative, the trig part of the problem where he figures out the angle wouldn't have made much sense, as the angle wouldn't have been between 0 and 90 degrees.
• Shouldn't the formula for the average velocity be (final minus initial over 2)?
(1 vote)
• Yes, just as long as there's constant acceleration.
(1 vote)
• Is it wrong to specify the complementary angle and say by the side of the vertical?
(1 vote)
• Yes. First, what does "by the side of the vertical" tell you? Which side? The North-East side? The North-West side? etc... Second, why take an extra step to figure out the complement when you have the angle? You have the answer and then change it. Yes, technically it would be a correct angle, but why bother.
(1 vote)
• why in every video sal writes v=at instead of v=u+at ?
is everytime we calculate final velocity the initial is taken zero ?
(1 vote)
• He writes Δv = at which is the same as Vfinal - Vinitial = at
(1 vote)
• What does below the horizontal mean? Or rather what is it referring too?
(1 vote)
• He is referring to the direction in which you draw your vector. If you imagine a coordinate system with the tail of the yellow line at the origin, the vector will point in the positive x-direction and in the negative y-direction (quadrant 4). The "horizontal" is the x-axis, so the vector is pointing below the x-axis at 78.7 degrees.
(1 vote)

Video transcript

In the last video, I realized that I made a mistake near the end, but it's not a good one, because I copied down the wrong number. And then that led to the rest of the problem not having the right values. We figured out that the vertical component of our velocity earlier in the video is negative 26.03 meters per second. But then something happened in my brain, and it became negative 29.03. So this is not negative 29.03. This is negative 26.03 meters per second. So the total velocity, we should have 26.03 squared over here. And then if we evaluate that-- and I just did it on my calculator-- 26.03 squared plus 5.61. Actually, I did it wrong again. So it's 5.21. Let me redo it. So 26.03 squared plus 5.21 squared under the radical gives us 26.55. So this right over here should be 26.55 meters per second. So that is the magnitude of our final velocity. So that is the magnitude of our total final velocity. And the angle, since this is not 29.03, this is 26.03, we can say that the tangent of this angle theta-- this is this value right over here-- is equal to the length of this vector right over here, which is the opposite side 26.03 over the length of this vector here, which is 5.21. Or we could say, if we take the inverse tangent of both sides of this, that the angle is equal to the inverse tangent or the arctangent, of 26.03 over 5.21. We have the inverse tangent of 26.03 divided by 5.21 gives us roughly 78.7 degrees. So the angle here is 78.7 degrees. And it's below the horizontal. So we can say that the vector, the final velocity vector, has a magnitude of 26.55 meters per second at a direction of 78.7 degrees below the horizontal. So I hope I didn't confuse you too much with that mistake at the end of that video.