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Part B) only. A particle is projected from a point O on the ground, with speed u at an...

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roshan-rox | Valedictorian

Posted August 26, 2013 at 4:47 PM via web

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Part B) only.

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,

A)

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`

B)

Deduce that the particle cannot strike the screen if `d > (2u / g )* sin alpha* (u cos alpha- v)`

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted August 26, 2013 at 5:32 PM (Answer #1)

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A particle is projected from a point O on the ground, with speed u at an angle `alpha` to the horizontal. The downward acceleration of the particle is equal to g. 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.

The component of the particle's velocity in the vertical direction is `u*sin alpha` . The distance the particle is in motion is `(2*u*sin alpha)/g` . The horizontal distance traveled by the particle in this duration of time is equal to `u*cos alpha*((2*u*sin alpha)/g)` . The distance of the screen from the point O after time `(2*u*sin alpha)/g` is `((2*u*v*sin alpha)/g) + d` . For the particle to not strike the vertical screen, `((2*u*v*sin alpha)/g) + d > u*cos alpha*((2*u*sin alpha)/g)`

=> `d > u*cos alpha*((2*u*sin alpha)/g) - (2*u*v*sin alpha)/g`

=> `d > (2*u*sin alpha)/g*(u*cos alpha - v)`

The particle does not strike the screen if `d > (2*u*sin alpha)/g*(u*cos alpha - v)`

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