1.

A body dropped from top of tower falls through 60 m during the last 2 seconds of its fall. The height of tower is (g = 10m//s^(2))

Answer»

95 m
80 m
90 m
60 m

SOLUTION :U = O, VELOCITY attained by ball at t = 2 second
`v=u+a t=0+g(t-2)`
distance traveled by ball in last 2sis given by
`h_(1)=v t+(1)/(2) at^(2)`
`60=g(t-2)xx2+(1)/(2) g(4)=20(t-2)+20`
at 4S, HEIGHT of tower
`=(1)/(2)g t^(2)=(1)/(2)xx10xx 4^(2)=80m`


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