For the second one with friction, it would have been much easier to just say that the energy in the system is equal to delta mech plus delta thermal energy, set initial KE equal to zero, and solve for vf. My Vf equation looked like vf = sqrt(2g[H-Hjump-mukd]
my approach in Problem no.2 is simple as K1+U1+Wothers=K2+U2.. then you can get velocity at the moment on the end of the ramp. then use kinematic equation on projectile motion in getting the Hmax, then I end up a height of 4.75m.
K1 is zero, U1 is at the height 8m, Wother is the frictional force. K2 is the kinetic energy at the moment the ski leaves the ramp, U2 is the potential energy at height 2m (ramp). Just plug in all the given, then you can directly solve the velocity of the ski at the moment he leaves the ramp, then just do kinematics to solve for Hmax.
when we found v at the height 2.0m, how do we know the direction of v? how would we have known that the v is in line with the angle of the ramp? in future questions, what determines whether the v is a v_x or v_y or v of some ramp it's being launched? If the surface was flat, would v be in the x direction i presume?
@@SatyajitMahapatra-ss3kpbecause at Hmax the y component of velocity is =0 , while the x component of V is always constant (that's an important characteristic of the projectile motion)
If that were the case you would GAIN gravitational energy when you move down rather then lose gravitational energy which gets converted to kinetic energy
Beautiful problem. It's so nice when everything fits together so nicely!
I did not study for my exams but watched this before and this video helped me so much. thank you
Glad it helped!
So clear and well-explained. Appreciate these videos so much; thank you!
For the second one with friction, it would have been much easier to just say that the energy in the system is equal to delta mech plus delta thermal energy, set initial KE equal to zero, and solve for vf. My Vf equation looked like vf = sqrt(2g[H-Hjump-mukd]
my approach in Problem no.2 is simple as K1+U1+Wothers=K2+U2.. then you can get velocity at the moment on the end of the ramp. then use kinematic equation on projectile motion in getting the Hmax, then I end up a height of 4.75m.
K1 is zero, U1 is at the height 8m, Wother is the frictional force. K2 is the kinetic energy at the moment the ski leaves the ramp, U2 is the potential energy at height 2m (ramp). Just plug in all the given, then you can directly solve the velocity of the ski at the moment he leaves the ramp, then just do kinematics to solve for Hmax.
the kinematic equation i used is Vy²-Voy²=-2g(y-yo), where Vy=0 at the Maximum height of trajectory.
when we found v at the height 2.0m, how do we know the direction of v? how would we have known that the v is in line with the angle of the ramp? in future questions, what determines whether the v is a v_x or v_y or v of some ramp it's being launched? If the surface was flat, would v be in the x direction i presume?
how come when we're looking for Hmax we're using cosine? I would have thought to use sine and Vy instead
😢I'm also thinking that
yeah that sucks, yeah not explain all the things clearly , he just gave us the results ... not explatin them at all
@@SatyajitMahapatra-ss3kpbecause at Hmax the y component of velocity is =0 , while the x component of V is always constant (that's an important characteristic of the projectile motion)
now how in the heck do you do this problem in 3 mins on the exam for ap physics 1
Thank you so much!
You're welcome!
I will try to solve this tomorrow and then I will watch It.
Thank sir it helps a lot
For which standard do you guys, in your country or state study theese kind of problem age guess?
16/17/18
14/15/16
Sir, Why is the value of gravity is not equal to negative -9.8m/s^2?
If that were the case you would GAIN gravitational energy when you move down rather then lose gravitational energy which gets converted to kinetic energy
Nding