Hi sir i found your videos very useful. The only thing in this video that got me confused is that are you assuming that the final velocity(V) value 14.7 is the same as the intial velocity(U) value 14.7 because in the question it doesn't say that it is constant speed and i am wondering can u always take the same value for V as U even if it doesn't say constant speed in a case like this?.
@@drongojuice6393 no its not wrong since with constant acceleration (and no air resistance) the velocity will always be the same when displacement is 0 To prove that anyways Find maximum height which in this case is 11.025 and then solve for the final velocity since initial velovity is 0 and u will find that it equals ± 14.7
ik this is old but like how do you know it's being thrown vertically upwards?? it doesn't specify it in the question i thought it meant like diagonally upwards
Hi Sir hope you keeping well Loved the vid. I am just wondering why did you use S=ut+1/2at^2 when you could have used something nicer like V=u +at???????????????
Very helpful video as always (: But I still struggle on knowing which velocity to choose as final velocity V and initial velocity U. Any tips on knowing which one to pick?? I have this issue for horizontal problems and vertical problems.
Always choose the velocity at the beginning of the projection (the one that it started with) as the initial velocity. The final velocity is the velocity at some time after the start that you want to calculate.
Hello there, how is possible that the result of the final problem is positive? ruclips.net/video/r604qVfIo6w/видео.html -14.7^2 / 2*(-9.8) = -11.02 is that because the displacement is - 11.02 and the height is actually the magnitude of it? Thank you, Petr.
Thanks it helped
How come final Velocity be 14.7 when it touches the ground..I didn't understand that
G.O.A.T!
I love the lessons
Hi sir i found your videos very useful. The only thing in this video that got me confused is that are you assuming that the final velocity(V) value 14.7 is the same as the intial velocity(U) value 14.7 because in the question it doesn't say that it is constant speed and i am wondering can u always take the same value for V as U even if it doesn't say constant speed in a case like this?.
Constant acceleration due to gravity
Gore rino ndichapasa! Thank you
Hi there. Why does it come down with the same velocity?
Because gravity works to slow things down *and* speed things up.
because the flight is symetrical
I know this is an old video but could you possibly explain why you can assume that the final velocity is -14.7?
It's not, dont listen to this video if you're doing as maths, its wrong.
@@drongojuice6393 no its not wrong since with constant acceleration (and no air resistance) the velocity will always be the same when displacement is 0
To prove that anyways
Find maximum height which in this case is 11.025 and then solve for the final velocity since initial velovity is 0 and u will find that it equals ± 14.7
you still haven't explained why we use -10 ms 2 despite its falling
ik this is old but like how do you know it's being thrown vertically upwards?? it doesn't specify it in the question i thought it meant like diagonally upwards
It says upwards
just a question, why at the start of the video were you saying we can take v to be -14.7 but then for part c you decided v is now 0.
V for total flight is -14.7
V for maximum height is 0 as the object comes to instantaneous rest
Two different things mate
Focus on what hes saying
If im taking the positive direction to be up for the first stage can i take the positive to be down for the second stage ?
Yes as long as you obey that sign convention with all your variables on the way down.
Hi Sir hope you keeping well Loved the vid. I am just wondering why did you use S=ut+1/2at^2 when you could have used something nicer like V=u +at???????????????
2yrs no answer it’s an evil world we live in
3 years no answer, this a diabolical world we live in
Very helpful video as always (:
But I still struggle on knowing which velocity to choose as final velocity V and initial velocity U. Any tips on knowing which one to pick?? I have this issue for horizontal problems and vertical problems.
Always choose the velocity at the beginning of the projection (the one that it started with) as the initial velocity. The final velocity is the velocity at some time after the start that you want to calculate.
Hello there,
how is possible that the result of the final problem is positive?
ruclips.net/video/r604qVfIo6w/видео.html
-14.7^2 / 2*(-9.8) = -11.02
is that because the displacement is - 11.02 and the height is actually the magnitude of it?
Thank you, Petr.
+Petr Stejskal It is not (-14.7)^2 but - 1 times 14.7^2. There are no brackets around the -14.7
Waw