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:

Question

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:

Infinity Learn · Accepted Answer

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$$ θ$$0where$$ θ$$0$$ $$=$$ angle of projectionθ$$0$$ $$=$$ $$tan$$$$-1(a)

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