A particle is dropped vertically downward under gravity. Consider the downward direction as positive. Ball collides elastically with ground.
Column-I | Column-II |
(ii) Velocity of particle changes with time as | [B] |
Text Solution
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Ans.
(i) [D]
(ii) [C]
(iii) [B]
(iv) [A]
Sol. v y = u y + a y t
y = u y t + 
a y t 2
* Velocity : u y = 0
v y = a y t (before collision)
v 0 = v y = gt 0 (at t = t 0 )
Which is straight line with positive slope.
v y = – v 0 + gt : after collision
* Displacement :
y = 
; before collision
y = – v 0 t + 
gt 2 ; after collision
Which is a parabola opening upwards.
* Distance-time graph is always increasing.
* Acceleration is constant and is equal to acceleration due to gravity.
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