A body falling from a vertical height of 0.5 m pierced through a distance of 20 cm in…
Physics · JEE Main · NTA Exams — Kinematics
A body falling from a vertical height of 0.5 m pierced through a distance of 20 cm in sand. It faces an average retardation in sand amounting to
[Take g = 10 m/s2]
- \(\text { — }\)10 m/s2
- \(\text { — }\)15 m/s2
- \(\text { — }\)25 m/s2
- \(\text { — }\)2.5 m/s2
Answer
(C) — 25 m/s 2
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