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## Class 11 Physics (India)

### Course: 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.

## 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.