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## Class 11 Physics (India)

### Course: Class 11 Physics (India)>Unit 8

Lesson 7: Projectiles launched from/to a height

# Total final velocity for projectile

Calculating the total final velocity for a projectile landing at a different altitude . Created by Sal Khan.

## Want to join the conversation?

• isn't the final vertical velocity is -26.03 instead of -29.03??
• Yes, Sal mixed them up. It should be 26.03, and the final degree measure is 78.7 degrees. There will be a correction in the next video.
• How would you do this using calculus?
• Can someone explain why the horizontal velocity haven't changed due to acceleration ? Why is acceleration 0 on the horizontal line ?
• Acceleration is zero on the horizontal because a few videos ago,we actually assume that drag is negligible. So, if drag is neglible and we ignore friction as well, there are no forces acting upon it. If there are no forces acting upon something then, the acceleration is zero.

This can be seen by f=ma (Newton's Second Law Of Motion)
So, if F= O, then it doesn't matter what mass is.
0= ma

The projectile has mass, but Acceleration will be zero because anything multiplied by zero is zero
m= O * accn
So, there is no acceleration, because there are no forces
• There is said:"we figured out... in the last video..." in this video. My question is: In what last video. There were another conditions in previous video "Launching and landing on different elevations".
• You can find the first part of the video on youtube:
• Why does Sal use the initial component of the velocity of 29.54m/s when it seems it should 0m/s from attaining its highest point? I thought the initial velocity would be from the apex of its trajectory since that is the immediate initial velocity before the projectile lands..
• We know the time that passed since the launch of the projectile, not the time that passed since it's highest point. So we're using the total change of time, from the very beginning, that's why we also use the launch velocity. You could calculate it from the apex, but then you have to first figure out the change of time from the apex to landing, which will be less than 5.67 s
• Two questions:
1. We will always take the final velocity vertically when calculating the total velocity?
2. Why did Sal not take the -26.03 value instead of +26.03??
• 1) Yes, even if it 0m/s. Although, if it were, we wouldn't really need to take it anyways.

2) 1-dimensional planes have only 2 directions (left, right, etc.) which are described using + and - symbols.

For 2 dimensions, it becomes complicated because there are multiple directions. Using + and - symbols would not work. So, directions are described using angles from the horizontal line.

But, regardless of this idea, or maybe because of it, Pythagoras' Theorem is used. And from the start, this theorem is only suited to find the magnitude of the vector. It is not suited to include directions in the calculation.

(Adding a negative sign presumably gives a false answer. Because in real life, even if we point the horizontal line in the opposite direction, the hypotenuse's length would be the same)

And finally, when the magnitude is obtained, it can be used alongside the angle/direction to describe the vector.
• At Sal calculates -26.026. When I subtract 9.8 from 29.54 and multiply that by 5.67, I get 111.93. Why do I not get -26.026? Is my math incorrect?
• For calculating the final vertical velocity, is it possible to use the formula: displacement=((initial velocity + final velocity)/2)*change in time? After substituting all the known values in, I get the result −26.017 which is only slightly different from Sal's result. Could this be because of small rounding errors or is this not the right equation to use?

Is the acceleration constant in this situation? does my equation include all the known and just one unknown variables.

Seems to me you got it right.

What you think??