Question

Two bodies are vertically thrown upward with initial velocity in the ratio 2:5. Find the ratio of maximum heights reached by them. Please quick in class

Answer 1

Given :

• Ratio of the initial velocity of two bodies which are thrown vertically upwards = 2 : 5

To find :

• Ratio of maximum heights reached by them

Solution :

Let the initial velocity of the first body be 2x and the initial velocity of the second body be 5x.

Calculating the maximum height reached by the first body :

Applying the third equation of motion :-

$\underline{\boxed{ \bold{v^2 - u^2 = 2aS}}}$

where,

• v denotes the final velocity
• u denotes the initial velocity
• a denotes the acceleration due to gravity
• S denotes the displacement/distance/height

We have,

• Initial velocity = 2x m/s

• Final velocity = 0 m/s

[The final velocity of the body when it is thrown upwards goes on decreasing and at the topmost point it is zero.]

• Acceleration due to gravity = - 10 m/s²

[The standard value of 'g' = 9.8 m/s² but after rounding it off we get 10 m/s² and we will take its value negative because when the body is thrown upwards then the acceleration due to gravity is negative and when it falls downward it's value is positive.]

Substituting the given values :

$\\ \twoheadrightarrow \quad\sf v^2 - u^2 = 2aS$

$\\ \twoheadrightarrow \quad\sf (0)^2 - (2x)^2 = 2(- 10)S$

$\\ \twoheadrightarrow \quad\sf - (2x)^2 = 2(- 10)S$

$\\ \twoheadrightarrow \quad\sf - 4x^{2} = - 20 \times S$

Transposing - 20 to the left hand side :

$\\ \twoheadrightarrow \quad\sf \dfrac{- 4x^{2}}{- 20} = S$

Cancelling the negative sign :

$\\ \twoheadrightarrow \quad\sf \dfrac{ \not4(x^{2})}{ \not20} = S$

$\\ \twoheadrightarrow \quad\sf \dfrac{ x^{2}}{5} = S$

$\\ \textsf{The maximum height reached by the first body = \dfrac{ \sf x^2}{ \sf 5} m}$

Now, calculating the maximum height reached by the second body :

Applying the third equation of motion :-

$\underline{\boxed{ \bold{v^2 - u^2 = 2aS}}}$

we have,

• Initial velocity = 5x m/s

• Final velocity = 0 m/s

• Acceleration due to gravity = - 10 m/s²

Substituting the given values :

$\\ \twoheadrightarrow \quad\sf v^2 - u^2 = 2aS$

$\\ \twoheadrightarrow \quad\sf (0)^2 - (5x)^2 = 2(- 10)S$

$\\ \twoheadrightarrow \quad\sf - (5x)^2 = 2(- 10)S$

$\\ \twoheadrightarrow \quad\sf - 25x^2 = - 20 \times S$

Transposing - 20 the left hand side :

$\\ \twoheadrightarrow \quad\sf \dfrac{- 25x^2}{- 20} = S$

Cancelling the negative sign :

$\\ \twoheadrightarrow \quad\sf \dfrac{ \not25(x^2)}{ \not20} = S$

$\\ \twoheadrightarrow \quad\sf \dfrac{ 5x^2}{4} = S$

$\\ \textsf{The maximum height reached by the second body = \dfrac{ \sf 5x^2}{ \sf 4} m}$

Now, we will calculate the ratio of the maximum height reached by the two bodies :

$\\ \dag \quad \sf Ratio = \dfrac{Maximum \: height \: reached \: by \: the \: first \: body}{Maximum \: height \: reached \: by \: the \: second \: body}$

Applying the values :

$\\ \twoheadrightarrow \quad \sf Ratio = \dfrac{ \dfrac{ {x}^{2} }{5} }{ \dfrac{ 5 {x}^{2} }{4}}$

$\\ \twoheadrightarrow \quad \sf Ratio = \dfrac{ {x}^{2} }{5} \div { \dfrac{ 5 {x}^{2} }{4} }$

$\\ \twoheadrightarrow \quad \sf Ratio = \dfrac{ {x}^{2} }{5} \times { \dfrac{ 4 }{ 5{x}^{2} } }$

$\\ \twoheadrightarrow \quad \sf Ratio = \dfrac{ 4{x}^{2} }{25 {x}^{2}}$

$\\ \twoheadrightarrow \quad \sf Ratio = \dfrac{ 4({x}^{2}) }{25 ({x}^{2})}$

$\\ \twoheadrightarrow \quad \sf Ratio = \dfrac{ 4( \: \!\!\!\not{x}^{2}) }{25 ( \: \!\!\not{x}^{2})}$

$\\ \twoheadrightarrow \quad \sf Ratio = \dfrac{ 4 }{25}$

Therefore, the ratio of the maximum height reached by both the bodies = 4 : 25

Answer 2

$\begin{gathered}\frak{ \pink{Given : }}\sf{\;\;\; The \: ratio \: of\: the\: initial\: velocity\: of\: two\: bodies\: which\: are\: thrown\: vertically\: upwards = \bf{2:5}}\end{gathered}$

$\begin{gathered}\frak{ \pink{To \: find : }}\sf{\;\;\; The\: ratio\: of\: maximum\: heights\: reached\: by\: them = \bf{?}}\end{gathered}$

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀

$\begin{gathered}\frak{ \pink{Solution : }}\end{gathered}$

As, we are given the ratio of the initial velocity of two bodies which are thrown vertically upwards is 2:5.

So, let the initial velocity of first body be "2x" and the initial velocity of second body be "5x".

$\pink{\underline{\sf For\: first \: body, \: we \: have} :}$

• Initial velocity, u = 2x m/s
• Final velocity, v = 0 m/s (because velocity at top point in air is 0)
• Acceleration due to gravity, a = 10 m/s²

And, we have to find height, S = ?

We know that :

• $\underline{\boxed{\bf v^2 - u^2 = 2aS}}$

$\sf : \implies (0)^2 - (2x)^2 = 2\times 10 \times S$

$\sf : \implies 0 - 4x^2 = 20 \times S$

$\sf : \implies x^2 = \cancel{\dfrac{20}{ - 4}} \times S$

$\sf : \implies x^2 = -5 \times S$

$\sf : \implies \dfrac{x^2}{-5} = S$

$\sf : \implies S = \dfrac{-x^2}{5}$

$\underline{ \sf \therefore The \: maximum \: height \: reached \: by \: the \: first \: body = \pink{\dfrac{ - x^2}{ 5} m}}$

$\pink{\underline{\sf For\: second \: body, \: we \: have} :}$

• Initial velocity, u = 5x m/s
• Final velocity, v = 0 m/s
• Acceleration due to gravity, a = 10 m/s²

And, we have to find height, S = ?

We know that :

$\underline{\boxed{\bf v^2 - u^2 = 2aS}}$

$\sf : \implies (0)^2 - (5x)^2 = 2\times 10 \times S$

$\sf : \implies 0 - 25x^2 = 20 \times S$

$\sf : \implies x^2 = \cancel{\dfrac{20}{ - 25}} \times S$

$\sf : \implies x^2 = \dfrac{-4}{5} \times S$

$\sf : \implies x^2 \dfrac{-5}{4} = S$

$\sf : \implies S = \dfrac{-5x^2}{4}$

$\underline{ \sf \therefore The \: maximum \: height \: reached \: by \: the \: first \: body = \pink{\dfrac{ - 5x^2}{ 4} m}}$

$\pink{\underline{\sf For \: Ratio} :}$

• $\sf Ratio = \dfrac{Maximum \: height \: reached \: by \: the \: first \: body}{Maximum \: height \: reached \: by \: the \: second \: body}$

$\sf : \implies Ratio = \dfrac{ \dfrac{ - {x}^{2} }{5} }{ \dfrac{ - 5 {x}^{2} }{4}}$

$\sf : \implies Ratio = \dfrac{ - {x}^{2} }{5} \times { \dfrac{ 4 }{ - 5{x}^{2} }}$

$\sf : \implies Ratio = \cancel{\dfrac{ - {4x}^{2} }{ - 25{x}^{2}}}$

$\sf : \implies Ratio = \dfrac{ 4}{25}$

$\: \: \: \: \: \: \:\:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \pink{\underline{\boxed{\pmb{\frak{Ratio = \dfrac{ 4}{25}}}}}}\:\bigstar$

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀

$\underline{ \bf Hence, \: the\: ratio \: of\: the \: maximum \: height\: reached\: by \: both\: the \: bodies \: is \: \pink{4 : 25}}.$