A stone is dropped from the …
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Payal Singh 6 years, 8 months ago
Distance travelled in last second of journey = 24.5 m
Applying,
sn = {tex}u+{1\over 2}a(2n-1){/tex}
as initial velocity u = 0 and body is under freely Fall so a = g = 9.8
{tex}24.1 = 0 +{9.8\over 2}(2n-1){/tex}
{tex}=> 5 = 2n-1{/tex}
=> n = 3
It means body was in air for 3 seconds.
Total distance covered in 3 seconds
{tex}s = {1\over 2}gt^2{/tex}
{tex}=> s = {1\over 2}\times 9.8\times 9{/tex}
=> s = 44.1 m
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