Question

a stone is thrown upwards with the velocity v at the half height its velocity will be?​

Answer 1

Explanation:

Given : -

If α and β are the zeroes of the polynomial x² + 12x + 4 = 0 .

Required to find : -

find the value of ( 1/α + 1/β ) ?

Solution : -

Quadratic equation : x² + 12x + 4 = 0

α and β are the zeroes of the polynomial .

Here we can solve this question using 2 methods !

The standard form of the Quadratic equation is

ax² + bx + c = 0

On comparing the standard form of a quadratic equation with the given polynomial

Here,

a = 1

b = 12

c = 4

1st method

We know that ;

There is a relationship between the zeroes of the Quadratic equation with respective to coefficients of the quadratic equation .

So,

The relation between the sum of the zeroes and the coefficients is ;

α + β = - coefficient of x/coefficient of x²

α + β = - b/a

α + β = -(12)/1

α + β = - 12

Similarly,

The relation between the product of the zeroes and the coefficients is ;

α.β = constant term/coefficient of x²

α.β = c/a

α.β = 4/1

α.β = 4

Now,

Let's find the value of ( 1/α + 1/β )

1/α + 1/β =

α + β/α.β

substituting the values ;

- 12/4

-3

2nd method

The Quadratic formula is ;

\boxed{ \sf{x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }}x=2a−b±b2−4ac

Here,

a = 1

b = 12

c = 4

Substituting these in the formula ;

\begin{gathered}\sf x = \dfrac{ - ( 12 ) \pm \sqrt{ ( 12 )^2 - 4 ( 1 ) ( 4)} }{ 2 ( 1 )} \\ \\ \\ \sf x = \dfrac{ - 12 \pm \sqrt{ ( 12 )^2 - 4 ( 1 ) ( 4)} }{ 2 } \\ \\ \\ \sf x = \dfrac{ - 12 \pm \sqrt{ 144 - 16} }{ 2 } \\ \\ \\ \sf x = \dfrac{ - 12 + \sqrt{128}}{2 } \quad ( and ) \quad x = \dfrac{ - 12 - \sqrt{128}}{2} \\ \\ \\ \implies \sf \alpha = \dfrac{ - 12 + \sqrt{128}}{2 } \quad ( and ) \quad \beta = \dfrac{ - 12 - \sqrt{128}}{2}\end{gathered}x=2(1)−(12)±(12)2−4(1)(4)x=2−12±(12)2−4(1)(4)x=2−12±144−16x=2−12+128(and)x=2−12−128⟹α=2−12+128(and)β=2−12−128

Now,

Let's find the value of 1/α + 1/β !

\begin{gathered}\sf \to \dfrac{1}{ \alpha} + \dfrac{1}{ \beta} \\ \\ \to \sf \dfrac{ \alpha + \beta }{ \alpha \beta } \\ \\ \to \sf \dfrac{ \dfrac{-12 + \sqrt{128}}{2} + \dfrac{- 12 - \sqrt{128}}{2} }{ \dfrac{- 12 + \sqrt{128}}{2} \times \dfrac{- 12 - \sqrt{ 128}}{2} } \\ \\ \\ \to \sf \dfrac{ \dfrac{- 12 + \sqrt{128} - 12 - \sqrt{128}}{2} }{\dfrac{( - 12 )^2 - ( \sqrt{128} )^2}{4} } \\ \\ \\ \to \sf \dfrac{ \dfrac{ - 12 - 12}{2 } }{ \dfrac{144 - 128}{4} } \\ \\ \\ \to \sf \dfrac{ \dfrac{- 24}{2} }{ \dfrac{16}{4} } \\ \\ \to \sf \frac{ - 24 }{ \dfrac{16}{2} } \\ \\ \to \sf \dfrac{ - 48 }{16} \\ \\ \implies \sf - 3\end{gathered}→α1+β1→αβα+β→2−12+128×2−12−1282−12+128+2−12−128→4(−12)2−(128)

Answer 2

Answer:

Let the velocity with which the stone was thrown be 'u'.

then the acceleration will be acceleration due to gravity ,i.e. '-g'.

height = $\frac{h}{2}$h/2

then by third equation of motion :

$v^{2}$v^2= u^2$u^{2}$ + 2gh

$v^{2}$v^2 = u^2 $u^{2}$ + 2(-g) $\frac{h}{2}$.

$v^{2}$ =  $u^{2}$ -2g $\frac{h}{2}$

$v^{2}$ =  $u^{2}$ - 2*10*$\frac{h}{2}$

$v^{2}$ =   $u^{2}$ - 10h

v = $\sqrt{u^{2}-10h }$.