Question

Two balls of different masses(one lighter and one heavier
) are thrown vertically upward with same initial speed which one of them have greater height

Answer 1

Both have the same height since it does not depend on the mass of the body thrown upwards. Let's prove it.

Let the masses of the heavier and lighter balls be M and m respectively and the initial velocity be u.

Here the height of the balls will be the displacement. Let the heights of the heaviest and the lightest balls be H and h respectively.

Here the acceleration of the balls is a = - g, since both are thrown vertically upwards.

We know that a body thrown vertically upwards has zero velocity at the maximum height, i.e., final velocity, v = 0.

Then by third kinematic equation,

v² = u² + 2as

0 = u² - 2gh

h = u² / 2g

Here we can see that the height only depends on the initial velocity and acceleration due to gravity but not on mass. Thus we can say that the heights are equal even if masses vary.

Well there's no need to have such a solution since none of the kinematic equations depend on the mass of the body! But we can clearly prove the statement by the law of conservation of mechanical energy too.

At the surface of the earth, the body has no potential energy since there's no height, and it has no kinetic energy at the maximum height since the velocity v = 0 there. Thus, in the case of heavier body,

$\dfrac {1}{2}Mu^2+Mg(0)=\dfrac {1}{2}Mv^2+MgH\\\\\\\dfrac {1}{2}Mu^2=MgH\\\\\\\dfrac {1}{2}u^2=gH\quad\longrightarrow\quad(1)$

and in the case of lighter body,

$\dfrac {1}{2}mu^2+mg(0)=\dfrac {1}{2}mv^2+mgh\\\\\\\dfrac {1}{2}mu^2=mgh\\\\\\\dfrac {1}{2}u^2=gh\quad\longrightarrow\quad(2)$

From (1) and (2), we get,

$H=h=\dfrac {u^2}{2g}$

which had been visited earlier.

Thus proved that the height does not depend on the mass of the body.

#answerwithquality

#BAL