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## Class 11 Physics (India)

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

Lesson 7: Projectiles launched from/to a height

# Launching and landing on different elevations

More complicated example involving launching and landing at different elevations. Created by Sal Khan.

## Want to join the conversation?

• I understand that time can't be negative for the purposes of this video, but can it be negative in any other situation? Please specify an example.
thanks •   The "time" in these videos is always the "change in time". So yes, it is quite logical to have a negative change in time. It would be before some event, such as before the projectile was launched.
• My only issue with this video is that you didn't specify the length of the 'well' in the middle. How do I know that the projectile won't land there instead of the second plateau? •  In other words, we are told that the projectile lands on the second plateau. That is a fact. What we could ultimately answer is the magnitude of the horizontal displacement (i.e. the distance from the cannon to the place where the projectile lands). The length of the 'well' will remain forever a mystery!
• @, why didn't Sal use the formula: s = Vi(x) + a(deltaT)
is it because the acceleration -9.8m/s^2 only applies in the vertical direction?

I'm just confused because I thought V(x) of 90cos53 was only for Vi(x) and not Vavg(x) • Sal, worked this problem in such a way to avoid using several other algebraic equations. He knows that Juniors and Seniors in high school and college know the quadratic equation. So he solved for the time by this means. There is another way (using basic algebra....takes forever)...For Y(direction)....You will need to use....Vf=Vi+at (Vf=0 Vi=90sin53 a=-9.8) t=7.33(this is just the time up) but total air time would be (7.3344*2= 14.67s)..to land on its original horizontal axis...and a=-9.8 because the object is going up and gravity is slowing its upward velocity...to where Vf=0 at 7.3344s)....Now as the objects falls, since the y-direction changes...its as if Vi=0 and Vf=90sin53...as it returns to its horizontal reference...BUT once it falls below it....we have a new calculation....to consider....Our Vi=90sin53 because it has an initial velocity but we now do not know its new final velocity...We have to find it.... Vf^2=Vi^2+2ad (Vi=90sin53 and a=9.8 and d=16) so Vf=74.03m/s..Now its is time to find how long it took to fall this 16m.......then Vf=Vi+at (Vf=74.03 Vi=71.88 a=9.8)and a=+9.8 because gravity is helping its Velocity increase....thus....t=0.22s (time it takes to fall the 16m)....final time in the air will be 14.67s+0.22= 14.89s....then to get the X-distance just use D=V*t...(D=90cos53*14.89= 806.5m....Sorry for the long winded explanation.
• Very stupid question:
Is the horizontal component of a velocity vector, how far the object was displaced or the total time it was in the air? • how come all the equations used are only applied to the vertical velocity of the cannonball? • hi, so does this mean that time up = time down also applies here when the elevation is different? sorry and thanks! • Sal has solved this using Quadratic formula, but I don't understand it is there any other way solve it?
Like by factoring or something? • hello . i want to ask why did sal put 90sin53 as a initial velocity instead of 90m/s , can someone explain it more?   