Question

A ball thrown upward from the top of tower with speed V reaches the ground in T1 second. If this ball is thrown downward from the top of the same tower with speed V it reaches the ground in T2 second. In what time the ball shall reach the ground if it is allowed to fall freely under gravity from top of the tower?

(a) T1+T2/2

(b) T2-T1/2

(c) √T1T2

(d) T1+T2

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Answer 1

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Answer 2

Answer:

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Explanation:

GivenQuestion−

Find the values of a and b for which the simultaneous linear equations x + 2y = 1 and (a - b)x + (a + b)y = a + b - 2 have infinitely many solutions.

\large\underline{\sf{Solution-}}Solution−

Given pair of linear equation is

\rm :\longmapsto\:x + 2y = 1 - - - (1):⟼x+2y=1−−−(1)

and

\rm :\longmapsto\:(a - b)x + (a + b)y = a + b - 2 - - - (2):⟼(a−b)x+(a+b)y=a+b−2−−−(2)

Now,

Comparing the given two equations with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we get  

a₁ = 1

b₁ = 2

c₁ = 1

a₂ = a - b

b₂ = a + b

c₂ = a + b - 2

Now, we know that,

System of two equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 have infinitely many solutions iff  

\boxed{ \bf{ \: \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}}}a2a1=b2b1=c2c1

So, on substituting the values, we get

\rm :\longmapsto\:\dfrac{1}{a - b} = \dfrac{2}{a + b} = \dfrac{1}{a + b - 2}:⟼a−b1=a+b2=a+b−21

On taking first and second member, we get

\rm :\longmapsto\:\dfrac{1}{a - b} = \dfrac{2}{a + b}:⟼a−b1=a+b2

\rm :\longmapsto\:a + b = 2(a - b):⟼a+b=2(a−b)

\rm :\longmapsto\:a + b = 2a - 2b:⟼a+b=2a−2b

\rm :\longmapsto\:a - 2a = - b- 2b:⟼a−2a=−b−2b

\rm :\longmapsto\: - a = - 3b:⟼−a=−3b

\bf :\longmapsto\: a = 3b - - - - - (1):⟼a=3b−−−−−(1)

Now, On taking first and third member, we get

\rm :\longmapsto\:\dfrac{1}{a - b} = \dfrac{1}{a + b - 2}:⟼a−b1=a+b−21

\rm :\longmapsto\:a + b - 2 = a - b:⟼a+b−2=a−b

\rm :\longmapsto \: b - 2 = - b:⟼b−2=−b

\rm :\longmapsto \: b + b = 2:⟼b+b=2

\rm :\longmapsto \: 2b= 2:⟼2b=2

\bf :\longmapsto \: b= 1:⟼b=1

On substituting the value of b in equation (1), we get

\bf :\longmapsto\:a = 3 \times 1 = 3:⟼a=3×1=3

\begin{gathered} \red{\begin{gathered}\begin{gathered}\rm :\longmapsto\:\bf\: Hence-\begin{cases} &\sf{a \: = \: 3} \\ \\ &\sf{b \: = \: 1} \end{cases}\end{gathered}\end{gathered}}\end{gathered}:⟼Hence−⎩⎪⎪⎨⎪⎪⎧a=3b=1