Two objects X and Y are thrown upwards simultaneously with the same speed. The mass of X is greater than that of Y. The air exerts equal resistive force on two objects, then
• A
X will be higher than Y
• B
Y will higher than X
• C
The two objects will reach the same height
• D
cannot say

Solution

Let the mass of the object A be $m_x$ and of object B be $m_y$ where $m_x > m_y$. Let the resistive force applied by air on the objects be F. Let the height reached by object X be $H_x$ and by object Y be $H_y$ . Let the speed they are launched be v. The balls will reach maximum height when their Kinetic Energy becomes zero. Writing the required equation we have;
$\dfrac {1}{2} \times m_x \times v^2 - F \times H_x - m_xg H_x=0$
$\dfrac {1}{2} \times m_y \times v^2 - F \times H_y - m_yg H_y=0$
Simplifying above we have;
$H_x = \displaystyle \frac {\frac{1}{2} \times m_x \times v^2}{F + m_xg}$
$H_y = \displaystyle \frac {\frac{1}{2} \times m_y \times v^2}{F+ m_yg}$
Dividing above two we have;
$\displaystyle \dfrac {H_x}{H_y} = \displaystyle \dfrac {m_x \times(F + m_yg)}{m_y\times (F + m_xg)}$

$\displaystyle \dfrac {H_x}{H_y} = \displaystyle \dfrac {m_xF + m_x m_yg}{m_yF + m_y m_xg}$
Since ${m_x}>{m_y}$ we have ${H_x}>{H_y}$.