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Question

The trajectory of a projectile in a vertical plane is y=αx-βx2, where α and β are constants and x and y, are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection θ and the maximum height attained H are respectively given by


A

tan-1α, α24β

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B

tan-1β, α22β

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C

tan-1βα, α2β

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D

tan-1α, 4α2β

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Solution

The correct option is A

tan-1α, α24β


Step 1: Find maxima of given equation

Given, trajectory of the projectile is described by the equation, y=αx-βx2

When maximum height (H) is achieved, y in the above equation is also maximum because y denotes the vertical position of the projectile and x denotes its horizontal position.

At maximum height, dydx=0,
α-2βx=0x=α2β
So the projectile reaches the highest point at α2β. The corresponding height will be,
y=αα2β-βα2β2=α22β-α24β=α24β

Therefore, maximum height Hmax=α24β

Step 2: Apply formula for angle of curve at a point

The angle between a parabola and the x-axis at any point is given as,
θ=tan-1dydxx=x0
where θ is the required angle and x0 is the abscissa of the point on the parabola

At the angle of projection, x=0. Thus,
θ=tan-1α-2β·0θ=tan-1α

Therefore, the angle of projection, θ=tan-1α.

Hence, option A is correct.


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