Sir please answer the question which follows, An objects is to be dropped from an aeroplane so as to hit a target on the ground.lf the aeroplane is traveling horizontally at 80m/s at a height of 2000m, calculate 1.the position of the aeroplane when the object is released. 2.the time for the object to hit the target from instant of release.
You are great teacher of math because you explain so clear ! Not many of my past teachers for example explain why vectors are longer if they are a greater multiple of a given one , if that makes sense. In this case a vector was a " unit vector " ( magnitude 1 ) and the second vector 1.5 times that . Again great job wish I could take math classes at the college you teach 👍
To reduce time say if this was on a contest or exam that had time constraints , you could have used cah.. adjacent / hypotenuse on the original angle you found ( 49.78 )instead of using sine ( opposite side over hypotenuse) After you found the blue angle by subtracting 90 - 49.78 =40.12
i would like to request more examples related to relative velocity, interception and so on. because i can understand ur explanations very well fr every sgle exmple shown... so i really want to watch more abt relative velocity from you, sir.. :Dthank you very much :D
Great video, explained a lot but i noticed (i my be wrong) in part a, that when using the sine rule, the ambiguous case of sine applies, so the angle can also be 180- (what you got) This means there will be two possible answers
theres a easier way to do this (at least easier in my opinion) w/o law of cos. just take the initial velocity vector and break it up into components relative to the river banks (Vx= (1m/s)*Cos(60)=.5 m/s, Vy=(1m/s)*Sin(60))=.866 m/s. The x component adds with the velocity vector of the river. if we take east as positive that gives -.5m/s+1.5m/s=1m/s. the y component of the initial velocity is unaffected by the river. vector addition of the components gives the same answer.
esha sharma for this you have to think logically as the wind is pushing towards the right it means the boat should be steered to the left to find the resultant direction towards destination
For part b) his original velocity is 1m/s but this is at 60 degrees. So it has a vertical and horizontal component. We should only be interested in the vertical component as we just want to cross the river. That component is sine 60 or Sqrt3/2 . Then S=D/t so t=D/S . We take the D=20m and S ( vertical comp of velocity) Sqrt3/2 and we divide 20 / (Sqrt3/2) and we get 23sec
Hi Chala, I agree. These were the first few videos I made when I was experimenting with screen recording. They are done many years ago before Khan Academy. They were sitting around on the computer for a while because I thought they were not good enough for public viewing. I hope I can share more recent videos, definitely clearer!
![Blox Fruits Dragon Rework Update [Full Stream]](http://i.ytimg.com/vi/EqDAp8udhm0/mqdefault.jpg)
It says he can paddle 1m/s in still water but the water here inst still how can u assume they canoe at 1m/s then?
Sir please answer the question which follows, An objects is to be dropped from an aeroplane so as to hit a target on the ground.lf the aeroplane is traveling horizontally at 80m/s at a height of 2000m, calculate
1.the position of the aeroplane when the object is released.
2.the time for the object to hit the target from instant of release.
You are great teacher of math because you explain so clear ! Not many of my past teachers for example explain why vectors are longer if they are a greater multiple of a given one , if that makes sense. In this case a vector was a " unit vector " ( magnitude 1 ) and the second vector 1.5 times that .
Again great job wish I could take math classes at the college you teach 👍
To reduce time say if this was on a contest or exam that had time constraints , you could have used cah.. adjacent / hypotenuse on the original angle you found ( 49.78 )instead of using sine ( opposite side over hypotenuse)
After you found the blue angle by subtracting 90 - 49.78 =40.12
Thank you for your kind words! Glad that I can be of help!
Very well explained.Thank you
Thank you so much for this video fella u helped me understand this fully. Appteciate the time work put into the videos, keep it up
i would like to request more examples related to relative velocity, interception and so on. because i can understand ur explanations very well fr every sgle exmple shown... so i really want to watch more abt relative velocity from you, sir.. :Dthank you very much :D
Great video, explained a lot but i noticed (i my be wrong) in part a, that when using the sine rule, the ambiguous case of sine applies, so the angle can also be 180- (what you got) This means there will be two possible answers
I didn't get it.😵
Why you use a2+b2-2abcos?
That is the cosine rule. Are you familiar with that?
Thank you so much, this helped me a lot!
theres a easier way to do this (at least easier in my opinion) w/o law of cos. just take the initial velocity vector and break it up into components relative to the river banks (Vx= (1m/s)*Cos(60)=.5 m/s, Vy=(1m/s)*Sin(60))=.866 m/s. The x component adds with the velocity vector of the river. if we take east as positive that gives -.5m/s+1.5m/s=1m/s. the y component of the initial velocity is unaffected by the river. vector addition of the components gives the same answer.
Thanks bruh, you got a subscriber
Please can you explain why the 60 degree angle on the bank is on the left hand side and not the right hand side of the VCW?
esha sharma for this you have to think logically as the wind is pushing towards the right it means the boat should be steered to the left to find the resultant direction towards destination
Correct me if I'm wrong but shouldn't Vc be 1m/s and not Vcw?As the question states the the velocity of the canoeist in STILL WATER is 1m/s
speed in still water mean relative to not actual
For part b) his original velocity is 1m/s but this is at 60 degrees. So it has a vertical and horizontal component.
We should only be interested in the vertical component as we just want to cross the river.
That component is sine 60 or Sqrt3/2 . Then S=D/t so t=D/S . We take the D=20m and S ( vertical comp of velocity) Sqrt3/2 and we divide 20 / (Sqrt3/2) and we get 23sec
Thank you!
OMG THIS IS SO FUCKING HELPFUL THANK YOU!!!
Good explanation but too many ok :P :D
THANK YOU!!!
Thanks a lot sir
Good really
okay?
OKAY! Thanks !
i bet you are from singapore
I bet chinese 🇨🇳
Visibility not perfect
Hi Chala, I agree. These were the first few videos I made when I was experimenting with screen recording. They are done many years ago before Khan Academy. They were sitting around on the computer for a while because I thought they were not good enough for public viewing. I hope I can share more recent videos, definitely clearer!
Ok ok ok ok ok ok ok