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
#### 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 $$ {H_x}>{H_y} $$.

- AX will be higher than Y
- BY will higher than X
- CThe two objects will reach the same height
- Dcannot say

$$ \dfrac {1}{2} \times m_x \times v^2 - F \times H_x - m_xg H_x=0 $$

$$ \dfrac {1}{2} \times m_y \times v^2 - F \times H_y - m_yg H_y=0 $$

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 $$ {H_x}>{H_y} $$.