Plz find the first part in the Q.6 and give the value of the range of the projectile plz.
What’s the answer ??
Will pseudo force not act ??
Have taken the answer according to ground frame only nd only changed d coordinate system
Pseudo force is taken into account when u change reference frame
That’s the main thing buddy
I also did the same thing, but the matter being in the answer, its just u^2sin(2alpha)/gcos(theta).
I saw the solution, and in that, they didn’t take the acceleration in the horizontal direction with respect to the block, into account.
It’s a JEE question. I’m amazed that they did such a silly mistake.
It sud b taken into account as no reason of not doing so
Certainly that’s why I am amazed.
Wait we r wrong
U is taken wrt box so now from box gsina comp should vanish
With respect to box, gsin(alpha) acts upwards, which will support the motion of the projectile, as the box moves under gravity only.
Suppose u r in the box will u even encounter that the same projectile is getting acc. With gsina as u r urself acc. There??? As we dont displace perpendicular to the inc. Plane we can see that in that direction its getting retarded with gcosa
Okay if u feel so, then try the same question with gsin(theta) be an acceleration with which the box is being pulled on the incline.
Well all the time we talked of gcos(alpha), it’s actually gsin(theta) and gcos(theta).