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 $$maximumdydx=0$$ or a−$$2bx=0$$ or $$x=a2b$$ (i) $$ymax=aa2b$$−$$ba2b2=a24b$$ (ii) $$dydx=a$$−$$2bxan$$ initial position x $$=$$ $$0dydxx=0=a=$$$$tan$$θθ$$=$$$$tan$$−$$1(a)$$

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