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

Infinity Learn · Accepted Answer

$$y=ax-bx2$$; for height or y to be maximum:  $$dydx=0$$   or   $$a-2bx=0$$   or   $$x=a2b$$ $$ymax=aa2$$ $$b-ba2$$ $$b2=a24$$ b $$dydxx-0=a=$$$$tan$$θ$$0$$, where θ$$0=$$ angle of projection  θ$$0=$$$$tan$$$$-1(a)$$

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