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If the greatest height to which a man can throw a stone is h, then the greatest horizontal distance upto which he can throw the stone is 2h.

The angle of projection for a projectile motion whose range R is n times the maximum height is tan-1(4/n).

The time of flight T and the horizontal range R of a projectile are connected by the equation gT2=2Rtan⁡θ, where θ is the angle of projection.

A ball is thrown vertically up. Another ball is thrown at an angle θ with the vertical. Both of them remain in air for the same period of time, then the ratio of heights attained by the two balls is 1 : 1.

answer is A.

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Detailed Solution

(a) Hmax=u22g=h Rmax=u2g=2h(b) R=nH u2sin⁡2θg=nu2sin2⁡θ2g tan⁡θ=4n⇒θ=tan−1⁡4n(c)R=u2sin⁡2θgT=2usin⁡θg⇒Rtan⁡θ=12gt2(d) T=2uyg,H=uy22g T1:T2::1:1⇒H1:H2::1:1

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