# 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`

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### 1 Answer

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,

`ucosalpha*t=d+vt`

`rArr t=d/(ucosalpha-v)`

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,

`h=usinalpha*t-1/2g t^2`

Plugging in the value of `t` ,

`h=(usinalpha*d)/(ucosalpha-v)-1/2g*(d/(ucosalpha-v))^2`

`=(2usinalpha*d(ucosalpha-v)-gd^2)/(2(ucosalpha-v)^2)`

`rArr 2h(ucosalpha-v)^2=2usinalpha*d(ucosalpha-v)-gd^2`

`rArr gd^2-2usinalpha*d(ucosalpha-v)+2h(ucosalpha-v)^2=0`

Hence the proof.