23 question second part
Asked by SanskarAgarwal86
14th September 2018,
4:19 PM
Answered by Expert
Answer:
When the smooth body is released from a height, it has to travel the distance 40 cm to reach the bottom.
If v is the velocity at bottom, then we have, v2 = 2×g×H = 2×9.8×0.4 = 7.84, hence v = 2.8 m/s.
when the smooth body climbs the inclined plane, it is subjected to retardation ( g×sin30 ).
hence the maximum vertical height h, it reaches in inclined plane is given by, v2 = 2×g×sin30×h or h = 7.84/(2×9.8×0.5) = 0.8 m = 80 cm
Answered by Expert
14th September 2018,
4:46 PM
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