00:01
Hello students in this question we have to approve that it hit the plane when the velocity is under root of v square 12g at an angle 10 inverse of under root of 2g h upon v okay so first of all i will draw the diagram so in this direction we will have v is equal to u cos alpha right and this bullet will go something like this and this is the point where it is fired and this is the jet plane and here it will cross the jet plane.
00:33
So here it will be hated by jet plane.
00:35
So now this height is h and in this direction we will in this direction we will have u.
00:43
So in the vertical direction we will have u -sine alpha.
00:46
Okay so now kinetic energy.
00:50
So we know that s is equal to u -t plus half a t square right so this is half a t square okay so now we know that this s will be equal to u and this u will be in the vertical direction that is sine alpha and to t minus half of acceleration will be g okay t square so from here i get an equation that is half gt square minus u sine alpha t plus h is equal to zero so now uh we know that since t is real so we can say that b of this equation that is b square minus 4 ac will be greater than 0.
01:36
So from here i say that u square sine square alpha minus 4 into half of g of h.
01:46
This will be greater than 0.
01:49
So from here i get u square sine square alpha is greater than 2g h okay and similarly we know that in the horizontal direction we have u course alpha so so we can say that the angle, this h would be under root of u square minus v square, that is the vertical direction.
02:20
So in this equation, if i put at the place of sine alpha, i would get u square, u square minus v square upon u square is greater than 2gh.
02:33
So from here what do i get is u square is greater than 2gh plus v plus v square...