A particle is projected from ground making an angle of $$tan$$⁡−$$12$$ with horizontal. Find the angle between its velocity vector and horizontal ground when its height above the ground will be equal to half of the maximum height attained by it.

Question

A particle is projected from ground making an angle of $$tan$$⁡−$$12$$ with horizontal. Find the angle between its velocity vector and horizontal ground when its height above the ground will be equal to half of the maximum height attained by it.

Infinity Learn · Accepted Answer

Maximum height, $$H=u2sin$$⁡$$2$$α$$2gwhere$$ α is the angle of projection. At height $$H2$$, Vcos⁡θ$$=ucos$$⁡αand $$(Vsin$$⁡θ$$)2=(usin$$⁡α$$)2$$−$$2$$.g.$$H2$$⇒$$(Vsin$$⁡θ$$)2=u2sin$$⁡$$2$$α−$$g(u2sin$$⁡$$2$$α$$2g)=u2sin$$⁡$$2$$α$$2$$⇒Vsin⁡θ$$=usin$$⁡α$$/2$$∴ $$tan$$⁡θ$$=Vsin$$⁡θVcos⁡θ$$=$$$$tan$$⁡α$$2=22=1$$⇒α$$=45o$$

A particle is projected from ground making an angle of $\tan^{- 1} \sqrt{2}$ with horizontal. Find the angle between its velocity vector and horizontal ground when its height above the ground will be equal to half of the maximum height attained by it.

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A particle is projected from ground making an angle of tan⁡−12 with horizontal. Find the angle between its velocity vector and horizontal ground when its height above the ground will be equal to half of the maximum height attained by it.

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

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A particle is projected from ground making an angle of $\tan^{- 1} \sqrt{2}$ with horizontal. Find the angle between its velocity vector and horizontal ground when its height above the ground will be equal to half of the maximum height attained by it.