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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)]


6 years ago


Answers : (1)


Dear   zahid sharief,

The initial velocity v0

v_0=v_{0x}\mathbf{i} + v_{0y}\mathbf{j}

 if the angle θ0 is known:

v0x = v0cosθ0
v0y = v0sinθ0



The horizontal displacement(xx0 )




xx0 = (v0cosθ0)t.




equation of projectile


Time to reach the maximum height

t={v_{0y}\over g}

Range of projectile


Maximum height                                         

y_{max}={{v^2\sin^2\theta}\over 2g}



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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6 years ago

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