A stone is projected from level ground with speed u and ann at angle `theta` with horizontal. Somehow the acceleration due to gravity (g) becomes double (that is 2g) immediately after the stone reaches the maximum height and remains same thereafter. Assume direction of acceleration due to gravity always vertically downwards.
Q. The horizontal range of particle is

Options

A. `3/4 (u^(2) sin 2 theta)/g`
B. `(u^(2) sin 2 theta)/(2g) (1+1/(sqrt(2)))`
C. `(u^(2))/g sin 2 theta`
D. `(u^(2) sin 2 theta)/(2g)(2+1/(sqrt(2)))`

Answer

B

Explanation

The time taken to reach maximum height and maximum height are
`t=(u sin theta)/(g)` and `H=(u^(2) sin^(2) theta)/(2g)`
For remaining half, the time of flight is
`t'=sqrt((2H)/((2g)))=sqrt((u^(2)sin^(2) theta)/(2g^(2)))=(t)/(sqrt(2))`
Total time of flight is `t + t' = t (1 + (1)/(sqrt(2)))`
`T =(u sin theta)/(2g)(1+(1)/(sqrt(2)))`
Also horizontal range is `= u cos theta xx T`
`= (u^(2) sin 2u)/(2g)(1+(1)/(sqrt(2)))`
Let `u_(y)` and `v_(y)` be initial and final vertical components of velocity.
`:. y_(y2)=2gH` and `v_(y2)=4gH`
`:. v_(y)=sqrt(2)u_(y)`
Angle `(phi)` final velocity makes with horizontal is
`tan phi=(v_(y))/(u_(x))=sqrt(2)(u_(y))/(u_(x))=sqrt(2) tan theta`.