Grade 12th PassMechanics

Explain the above problem with detailed solution. I can't able to understand

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Balamohan

8 Years agoGrade 12th Pass

1 Answer

Arun

8 Years ago

Dear student

I have taken Vo = u

and

h = ( u ² / g ) cos ϕ [ sin ϕ tan α – cos ϕ ] … where … ϕ = θ + α …
If the projectile, moving with final velocity V (components Vx , Vy) hits the
inclined surface at an angle of 90°, then … Vx / Vy = tan α … since …
… V ⊥ inclined surface and Vy ⊥ x-axis … where … Vx = u cos ϕ = constant …
… while … Vy = ( u sin ϕ ) – g t … ϕ = angle of projection with respect to the
horizontal x-axis … now, assume that the projectile strikes the inclined surface
at a vertical height h above the point of projection, then … tan α = h / x
… which gives … x = h / tan α … and with the projectile fired at the origin of the
coordinate system at an angle θ with respect to the incline …
… x = u cos ( θ + α ) t = h / tan α …-->… t = h / [ u cos ϕ tan α ] … where …
ϕ = θ + α … it follows that … Vy = ( u sin ϕ ) – g h / [ u cos ϕ tan α ] … which can
be rewritten as … Vy = [ u ² sin ϕ cos ϕ tan α – g h ] / [ u cos ϕ tan α ] … so that …
… tan α = Vx / Vy = u cos ϕ / { [ u ² sin ϕ cos ϕ tan α – g h ] / [ u cos ϕ tan α ] }
……….. = [ u ² cos ² ϕ tan α ] / [ u ² sin ϕ cos ϕ tan α – g h ] … which reduces to …
… 1 = [ u ² cos ² ϕ ] / [ u ² sin ϕ cos ϕ tan α – g h ] …
… u ² sin ϕ cos ϕ tan α – g h = u ² cos ² ϕ … solving for h , we finally get …
… h = ( u ² / g ) cos ϕ [ sin ϕ tan α – cos ϕ ] … where … ϕ = θ + α …