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 angle `phi` which the velocity vector of stone makes with horizontal just before hitting the ground is given by:

Options

A. `tanphi=2tantheta`
B. `tanphi=2cottheta`
C. `tanphi=sqrt2tantheta`
D. `tanphi=sqrt2costheta`

Answer

C

Explanation

The time taken to reach maximum height and maximum height are
`t=(usintheta)/(g)` and `H=(u^2sin^2theta)/(2g)`
For remaining half, the time of flight is
`t^`=sqrt((2H)/((2g)))=sqrt((u^2sin^2theta)/(2g^2))=9t)/(sqrt2)`
Total time of flight is `t+t^`=t(1+(1)/(sqrt2))`
`T=(usintheta)/(g)(1+(1)/(sqrt2))`
Also horizontal range is
`=ucosthetaxxT=(u^2sin2theta)/(2g)(1+(1)/(sqrt2))`
Let `u_y` and `v_y` be initial and final vertical camponents of velocity.
`u_y^2=2gH` and `v_y^2=2gH`
`v_y=sqrt2u_y`
Angle `(phi)` final velocity makes with horizontal is
`tanphi=(v_y)/(u_x)=sqrt2(u_y_)/(u_x)=sqrt2tantheta` or `u_x=5`m.s
`u=sqrt(u_x^2+u_y^2)=sqrt(28)`