A body dropped from top of a tower falls through 40m during last two second of its motion of its fall. The height of tower in m is=?
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Hello......
Friend.......
Your ans is....
Ans. suppose that during travelling 40m in last 2 sec its initial velocity was u.
Explain...
ut+1/2gt^2=40 ( for last 2 sec)
2u-19.6=40
u=29.8m/s
now the considering from top to this position okey.
u=0 and v=29.8m/s
then by v^2-u^2=2gh
29.8^2=2*9.8*h
h=45.3m
so finally total height of tower is
=45.3+40= 85.30m
Thankyou....
Friend.......
Your ans is....
Ans. suppose that during travelling 40m in last 2 sec its initial velocity was u.
Explain...
ut+1/2gt^2=40 ( for last 2 sec)
2u-19.6=40
u=29.8m/s
now the considering from top to this position okey.
u=0 and v=29.8m/s
then by v^2-u^2=2gh
29.8^2=2*9.8*h
h=45.3m
so finally total height of tower is
=45.3+40= 85.30m
Thankyou....
anujana:
10
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