Question

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

Answer 1

Answer:

4:25

Explanation:

As per the provided information in the given question, we have :

• Two bodies are vertically thrown upward with initial velocity in the ratio 2:5.

We are asked to calculate the ratio of maximum heights reached by them.

Let us suppose their initial velocities as 2x and 5x.

$\longmapsto \: \boxed{\sf { Ratio_{(Heights)}= \dfrac{Height \; reached_{(1st \; body)} }{Height \; reached_{(2nd \; body)}} }}$

★ Calculating height reached by the 1st body :

By using the third equation of motion,

$\longrightarrow \boxed{\sf {v^2 - u^2 = 2gh}}$

Whenever the body is thrown vertically upwards, its final velocity (v) is 0.

• Final velocity (v) = 0
• Initial velocity (u) = 2x
• Acceleration due to gravity (g) = -10 m/s²
• Height ($\sf h_1$) = ?

$\longrightarrow \sf { (0)^2 - (2x)^2 = 2 \times -10 \times h_1}$

$\longrightarrow \sf { 0 - 4x^2 = -20 \times h_1}$

$\longrightarrow \sf { -4x^2 = -20 \times h_1}$

$\longrightarrow \sf { \cancel{\dfrac{-4x^2}{-20}} = h_1}$

$\longrightarrow \sf { \dfrac{1x^2}{5} = h_1} \dots \mathfrak { Equation \;1 }$

★ Calculating height reached by the 2nd body :

By using the third equation of motion,

$\longrightarrow \boxed{\sf {v^2 - u^2 = 2gh}}$

• Final velocity (v) = 0
• Initial velocity (u) = 5x
• Acceleration due to gravity (g) = -10 m/s²
• Height ($\sf h_2$) = ?

$\longrightarrow \sf { (0)^2 - (5x)^2 = 2 \times -10 \times h_2}$

$\longrightarrow \sf { 0 - 25x^2 = -20 \times h_2}$

$\longrightarrow \sf { -25x^2 = -20 \times h_2}$

$\longrightarrow \sf { \cancel{\dfrac{-25x^2}{-20}} = h_2}$

$\longrightarrow \sf { \dfrac{5x^2}{4} = h_2} \dots \mathfrak { Equation \; 2}$

★ Calculating the ratio of maximum heights reached by them :

We know ,

$\longmapsto \: \boxed{\sf { Ratio_{(Heights)}= \dfrac{Height \; reached_{(1st \; body)} }{Height \; reached_{(2nd \; body)}} }}$

Substituting the value of equation 1 and equation 2.

$\longrightarrow \sf {Ratio_{(Heights)}= \dfrac{\cfrac{1x^2}{5} }{ \cfrac{5x^2}{4}} }$

$\longrightarrow \sf {Ratio_{(Heights)}= \dfrac{1x^2}{5} \times \dfrac{4}{5x^2} }$

$\longrightarrow \sf {Ratio_{(Heights)}=\dfrac{4x^2}{25x^2} }$

$\longrightarrow \sf {Ratio_{(Heights)}= \dfrac{4}{25} }$

$\longrightarrow \underline{\boxed{\sf {Ratio_{(Heights)}= 4\ratio 25 }}} \; \bigstar$

Therefore, 4:25 is the required answer.