In the problem 150, the time taken to return to the ground from the maximum height:

Options

A. is almost same as in the absence of friction
B. decreases by 1%
C. increases by 1%
D. increases by 11%

Answer

a

Explanation

Here, `H'_(m)=(u^(2)sin^(2) theta)/(2xx(11g)/10)=(10u^(2)sin^(2) theta)/(22g)`
Using `x=ut+1/2at^(2) where x=H'_(m), u=0` and
`a=g-g/10=(9g)/10` , we find `t=sqrt(H'_(m)//a)`
`=sqrt(200/198)xx(u sin theta)/g`
It is almost equal to the time of fall in the absence of friction.