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
a2b, tan-1(2b)
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 maximum,dydx = 0 or a-bx = 0or x = a2b(i) ymax. = a(a2b)-b(a2b)2 = a24b(ii) (dydx)x= 0 = a = tan θ0where θ0 = angle of projectionθ0 = tan-1(a)