Q.

The trajectory of a projectile in a vertical plane is y=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

b24a, tan−1(b)

c

a24b,tan−1(a)

d

2a2b, tan−1(a)

answer is C.

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Detailed Solution

y=ax−bx2For height or y to be maximumdydx=0 or a−2bx=0 or x=a2b (i) ymax=aa2b−ba2b2=a24b (ii) dydx=a−2bxan initial position x = 0dydxx=0=a=tanθθ=tan−1(a)
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