Two masses $m_{1} a n d m_{2} (m_{1} > m_{2})$ are falling from the same height when same air resistance acts on them

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$m_{1}$ has more velocity on striking the ground

all the above

$m_{1}$ has more acceleration

$m_{1}$ reaches the ground early

answer is D.

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Detailed Solution

Given that two masses $m_{1} a n d m_{2} (m_{1} > m_{2})$ are falling from the same height both are said to be experiencing same air friction f.

We can have net force on $m_{1}$

$F_{n e t} = m_{1} g - f m_{1} a_{1} = m_{1} g - f ∴ a_{1} = g - \frac{f}{m_{1}} \to (1) S i m i l a r l y n e t f o r c e o n m_{2} F_{n e t} = m_{2} g - f m_{2} a_{2} = m_{2} g - f a_{2} = g - \frac{f}{m_{2}} \to (2) \sin c e m_{1} > m_{2} f r o m (1) & (2) w e g e t a_{1} > a_{2} a n d h e a v i e r b o d i e s r e a c h e s g r o u n d e a r l i e r i . e . m_{1} w i l l r e a c h t h e g r o u n d e a r l i e r,, m_{1} w i l l g a i n g r e a t e r a c c e l e r a t i o n . O b v i o u s l y m_{1} w i l l s t r i k e t h e g r o u n d w i t h g r e a t e r v e l o c i t y .$