Question

16. The ratio of the heights from which two bodies are
dropped is 3 : 5, respectively. The ratio of their final
velocities is
(a) 15:13
(b) 13:15
(c) 9:25
(d) 5:3​

Answer 1

Answer:

Given:

Ratio of height from which 2 bodies have been dropped is 3:5.

To find:

Ratio of final velocities

Concept:

Since gravitational acceleration is constant for both the bodies , we can apply the Equations of Kinematics. Also we need to assume that gravitational acceleration in this case doesn't change with height.

Calculation:

Using 3rd Equation of Kinematics :

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

For dropping, we can say that initial velocity is 0 m/s ,

$\bigstar \: \: \boxed{ \sf{ \: \: {v}^{2} = 2gh}}$

Taking x as constant of proportionality :

So , ratio of final velocities will be :

${(v2)}^{2} : {(v1)}^{2} = \cancel{2g}(h2) : \cancel{2g}(h1)$

$= > {(v2)}^{2} : {(v1)}^{2} = (h2) : (h1)$

$= > v2 : v1 = \sqrt{h2} : \sqrt{h1}$

$= > v2 : v1 = \sqrt{3x} : \sqrt{5x}$

$= > v2 : v1 = \sqrt{3} : \sqrt{5}$

So final answer :

$\boxed{ \red{ \huge{ \sf{v2 : v1 = \sqrt{3} : \sqrt{5} }}}}$

Answer 2

SoluTion :

✴ Given :-

▪ The ratio of heights from which two bodies are dropped is 3:5 respectively.

✴ To Find :-

▪ Ratio of their final velocities

✴ Concept :-

✈ This question is completely based on concept of 'Energy conservation'.

✴ Calculation :-

$\dashrightarrow\sf\:\red{Initial\:potential\:energy=Final\:kinetic\:energy}\\ \\ \circ\bf\:potential\:energy=mgH\\ \\ \circ\bf\:kinetic\:energy=\dfrac{1}{2}mv^2\\ \\ \dashrightarrow\sf\:\dfrac{M_1gH_1}{M_2gH_2}=\dfrac{\frac{1}{2}M_1{v_1}^2}{\frac{1}{2}M_2{v_2}^2}\\ \\ \dashrightarrow\sf\:\dfrac{{v_1}^2}{{v_2}^2}=\dfrac{H_1}{H_2}\\ \\ \dashrightarrow\sf\:\dfrac{v_1}{v_2}=\sqrt{\dfrac{3}{5}}\\ \\ \dashrightarrow\:\boxed{\tt{\orange{\large{v_1:v_2=\sqrt{3}:\sqrt{5}}}}}$

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Note : you can also calculate this question by second equation of kinematics.