question_answer

A ball is dropped from the top of a building. The ball takes 0.5 s to fall past the 3 m length of a window some distance from the top of the building. If the velocities of the ball at the top and at the bottom of the window are \[{{v}_{T}}\] and \[{{v}_{B}}\] respectively, then \[(take\text{ }g=10\text{ }m/{{s}^{2}})\]

A) \[{{\text{v}}_{\text{T}}}\text{+}{{\text{v}}_{\text{B}}}\text{=12m}{{\text{s}}^{-1}}\]

B) \[{{\text{v}}_{\text{T}}}-{{\text{v}}_{\text{B}}}=4.9\text{m}{{\text{s}}^{-1}}\]

C) \[{{\text{v}}_{\text{B}}}{{\text{v}}_{\text{T}}}\text{=1m}{{\text{s}}^{-1}}\]

D) \[{{\text{v}}_{\text{B}}}\text{/}{{\text{v}}_{\text{T}}}\text{=1m}{{\text{s}}^{-1}}\]

Correct Answer: B

Solution :

\[\text{v}_{\text{B}}^{\text{2}}\text{=v}_{\text{T}}^{\text{2}}\text{+2}\left( \text{10} \right)\left( \text{3} \right)\] \[\Rightarrow \text{v}_{\text{B}}^{\text{2}}\text{=v}_{\text{T}}^{\text{2}}\text{+60                        }....\text{(i)}\] \[\text{Also, }{{\text{v}}_{B}}={{\text{v}}_{T}}+\left( 10 \right)\left( 0.5 \right)\text{       }....\text{(ii)}\] Solve eqs. (i) and (ii) We get \[{{\text{v}}_{T}}-{{\text{v}}_{B}}=4.9\text{ m/s}\]