Two Particles A & B are projected from the same point with the same velocity u of projection but at different angles (x) & (y), such that the maximum height of A is two-third of the horizontal range of B.Then which of the following relations are true ?

(a) Range of A = Maximum height of B

(b) Maximum horizontal range of A = u^2/g and this occurs when (y) = 1/2 sin^-1 3/8

(c) 3(1 - cos 2{x}) = 8 sin 2(y)

(d) Maximum value of (y) is 1/2 sin^-1 3/4

[Ans: (b) , (c) , (d)]

Dear   zahid sharief,

The initial velocity v₀

 if the angle θ₀ is known:

v_0x = v₀cosθ₀

v_0y = v₀sinθ₀

The horizontal displacement(x − x₀ )

x − x₀ = (v₀cosθ₀)t.

..

equation of projectile

Time to reach the maximum height

Range of projectile

Maximum height                                         

maimum height of  A     y1(max) =   v^2 (1-cos 2x)/4 *g       here  2 sin²x= 1-cos2x

range of                        r2(max) = v^2*sin(2y)/g     

compare them we get    3(1 - cos 2{x}) = 8 sin 2(y)

maimum value    (y)= 1/2 [sin^_-1 {3/8(1 - cos 2{x}) } ]

say cos2x=-1

we get maximum value   that is     max(y)= 1/2 [sin^_-1 {3*2/8} ]

                                   (y)= 1/2 [sin^_-1 3/4 ]

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