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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

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										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(x − x0 )

$x-x_0=v_{0x}t+\begin{matrix}\frac{1}{2}\end{matrix}at^2$

x − x0 = (v0cosθ0)t.
.$y-y_0=(v_0\sin\theta_0)t-\begin{matrix}\frac{1}{2}\end{matrix}gt^2$.

$y-y_0=(v_0\sin\theta_0)t-\begin{matrix}\frac{1}{2}\end{matrix}gt^2$

equation of projectile
$y=x\tan(\theta)-x^2g/(2v^2_{0}cos^2(\theta))$
Time to reach the maximum height
$t={v_{0y}\over g}$
Range of projectile
${r=\frac{v^2}{g}\sin(2\theta)}\,$
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 ]

All the best.
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Rathod Shankar
IIT bombay

6 years ago

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