a stone is dropped from rest from the top of a tower 19.6m high. find the distance covered during the last second of its motion ... please ans fast its urgent???
NobitaNobi6:
You have to use 2nd equation of motion.
S=ut+1/2at^2
it cant be used because question is asking about a particular second not a interval of time
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Listen,
s = ut + 1/2 at^2
Now, the the stone is dropped from rest, u =0
Let the distance covered in last second be x
Therefore
19.6 - x = 0*t + 1/2 *10 1^2
=19.6 - x = 5
Therefore, x = 14.6
s = ut + 1/2 at^2
Now, the the stone is dropped from rest, u =0
Let the distance covered in last second be x
Therefore
19.6 - x = 0*t + 1/2 *10 1^2
=19.6 - x = 5
Therefore, x = 14.6
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