The distance travelled by a freely falling body from rest in first 5 seconds of its motion is equal to the distance travelled in last second of its journey. The time of its fall is
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The distance covered in first 3 seconds is
S=12gt2S=12×9.8×(3)2S=45m
If the ball takes n seconds to to fall down,the distance covered in nth second is
S=12g(2n−1)S=10n−5
As distance covered in last second is equal to distance covered in first 3 sec
10n−5=45n=5
The total height is h=12gt2=102×25=125m
The time of its travel is
t=2hg‾‾‾√t=2×12510‾‾‾‾‾√t=5 sec
Regards.
S=12gt2S=12×9.8×(3)2S=45m
If the ball takes n seconds to to fall down,the distance covered in nth second is
S=12g(2n−1)S=10n−5
As distance covered in last second is equal to distance covered in first 3 sec
10n−5=45n=5
The total height is h=12gt2=102×25=125m
The time of its travel is
t=2hg‾‾‾√t=2×12510‾‾‾‾‾√t=5 sec
Regards.
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