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Q. A ball A is dropped from a building of height $ 45\, m $ . Simultaneously another identical ball $ B $ is thrown up with a speed $ 50 \,m \,s^{-1} $ . The relative speed of ball $ B $ w.r.t. ball A at any instant of time is (Take $ g = 10\, m \,s^{-2}) $

Motion in a Straight Line

Solution:

Here, $ u_A = -0 $ , $ u_B = +50\, m\, s^{-1} $
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$ a_{A}=-g $ , $ a_{B}=-g $
$ u_{BA}=u_{B}-u_{A}=50\,m\,s^{-1}-(-0)\,m\,s^{-1}=50\,m\,s^{-1} $
$ a_{BA}=a_{B}-a_{A}=-g-\left(-g\right)=0 $
$ \because v_{BA}=u_{BA}+a_{BA}t\quad\left(As\,a_{BA}=0\right) $
$ \therefore v_{BA}=u_{BA} $
As there is no acceleration of ball $ B $ w.r.t to ball $ A $ , therefore the relative speed of ball $ B $ w.r.t ball $ A $ at any instant of time remains constant $ ( = 50\, m\, s^{-1}) $ .