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The trajectory of a projectile in a vertical plane isy=ax−bx2 , where a and b are constants and x and y are respectively horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal are:

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a

b22a,tan−1(b)

b

a2b,tan−1(2a)

c

a24b,tan−1(a)

d

2a2b,tan−1(a)

answer is C.

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Detailed Solution

y=ax−bx2 For height or y to be maximum,dydx=0 or a−2bx=0 or x=a2b (i)ymax=a(a2b)−b(a2b)2=a24b (ii)(dydx)x=0=a=tanθ0 (where θ0=angle of projection) ∴ θ0=tan−1(a)

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The trajectory of a projectile in a vertical plane isy=ax−bx2 , where a and b are constants and x and y are respectively horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal are: