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zahid sharief Grade: 11
        


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)]


 


7 years ago

Answers : (1)

Rathod Shankar AskiitiansExpert-IITB
69 Points
										

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 )


 


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

 


xx0 = (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 ]


 


 


 




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Askiitians Expert


Rathod Shankar


IIT bombay

7 years ago
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