Question

two bodies are thrown vertically upwards with their initial velocities are in the ratio 4:3 ,then ratio of their maximum heights attained by them​

Answer 1

Explanation:

Hope this will help you

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$So velocity vectors of both are in two different directions.

The given ratio of velocity is

$\large{u_1 : u_2 = 4 : 3}$

$\huge{\bold{\underline{Explanation:-}}}$

From third kinematic equation,

$\large{\bold{v^2 - u^2 = 2as}}$

At heighest point the velocity "v" will be Zero.

Substituting the value,

$\large{0 - u^2 = - 2gH}$

As the ball is thrown upward the acceleration due to gravity is negative.

$\large{\boxed{H = \dfrac{u^2}{2g}}}$

Now,

Acceleration due to gravity will be constant.

Therefore,

$\large{\boxed{H \propto u^2}}$

(H is directly proportional to the square of initial velocity)

Now,

$\large{ \dfrac{H_1}{H_2} = \dfrac{(u_1)^2}{(u_2)^2}}$

$\large{ \dfrac{H_1}{H_2} = \left[ \dfrac{u_1}{u_2} \right]^2}$

Now, Substituting the ratio of velocities

$\large{ \dfrac{H_1}{H_2} = \left[ \dfrac{4}{3} \right]^2}$

$\large{ \dfrac{H_1}{H_2} = \dfrac{(4)^2}{(3)^2}}$

$\huge{\boxed{\boxed{ \dfrac{H_1}{H_2} = \dfrac{16}{9}}}}$

So, the ratio of heights is 16 : 9.