A particle is projected from a point O on the ground, with speed u at an angle `alpha` to the horizontal, under gravity. At the same instant, a vertical screen at perpendicular distance d from O...
A particle is projected from a point O on the ground, with speed u at an angle `alpha` to the horizontal, under gravity. At the same instant, a vertical screen at perpendicular distance d from O and at right angles to the vertical plane of motion of the particle, is made to move away from the particle, in a horizontal direction with uniform speed v.
If the particle strikes the screen at a height h above the ground, show that
`u cos alpha > v` , and `g(d^2) -2u sin alpha (u cos alpha - v ) d+2h(u cos alpha -v)^2 = 0`
Let the particle hits the screen after time `t` from being projected.
The particle will have a constant horizontal velocity of `ucosalpha` , and a decelerating vertical motion, having initial velocity of `usinalpha` .
The screen is moving away from the particle with a horizontal velocity `v` .
As the particle is beginning its flight at a distance `d` behind the screen, it is implied that the particle hits the screen only when its horizontal velocity is greater than that of the screen, i.e.
`ucosalphagtv` (refer to the attached image)
Consider the horizontal motion of both the particle as well as the screen, for hitting the screen,
Now consider the vertical motion of the particle, it hits the screen at a height h above the ground. From Newton's laws of motion,
Plugging in the value of `t` ,
Hence the proof.