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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)]


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)]


 


Grade:11

1 Answers

Rathod Shankar AskiitiansExpert-IITB
69 Points
13 years ago

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 ]

 

 

 


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Askiitians Expert

Rathod Shankar

IIT bombay

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