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

A stone is projected from level ground with speed u and ann at angle t

1.	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
Answer» `(3)/(4)(U^(2)sin2theta)/(g)` `(u^(2)sin 2theta)/(2g) (1+(1)/(SQRT(2)))` `(u^(2))/(g) sin 2theta` `(u^(2)sin2theta)/(2g) (2+(1)/(sqrt(2)))` SOLUTION :The time taken to REACH maximum HEIGHT and maximum height are `t=(usintheta)/(g)andH=(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=t(1+(1)/(sqrt(2)))` `T=(usintheta)/(g)=(1+(1)/(sqrt(2)))` So horizontal range is u `costhetaxxT=(u^(2)sin2theta)/(2g)(1+(1)/(sqrt(2)))`