Kinematics Question 420
Question: 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 $ (takeg=10m/s^{2}) $
Options:
A) $ {{v} _{T}}\text{+}{{v} _{B}}\text{=12m}{{s}^{-1}} $
B) $ {{v} _{T}}-{{v} _{B}}=4.9m{{s}^{-1}} $
C) $ {{v} _{B}}{{v} _{T}}\text{=1m}{{s}^{-1}} $
D) $ {{v} _{B}}\text{/}{{v} _{T}}\text{=1m}{{s}^{-1}} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ v _{B}^{2}\text{=v} _{T}^{2}\text{+2}( 10 )( 3 ) $
$ \Rightarrow v _{B}^{2}\text{=v} _{T}^{2}\text{+60 }….\text{(i)} $
$ \text{Also, }{{v} _{B}}={{v} _{T}}+( 10 )( 0.5 )\text{ }….\text{(ii)} $ Solve eqs. (i) and (ii) We get $ {{v} _{T}}-{{v} _{B}}=4.9\text{ m/s} $
BETA
