A particle is projected horizontally with a speed u from the top of plane inclined at an angle theta with the horizontal. How far from the point of projection will particle strike the plane?

A particle is projected horizontally with a speed u from the top of pl

1.	A particle is projected horizontally with a speed u from the top of plane inclined at an angle theta with the horizontal. How far from the point of projection will particle strike the plane?
Answer» Solution :Suppose particle is projected from height y, it strikes the ground at a horizontal distance x, then distance `R=sqrt(x^(2)+y^(2))`……..i It t is the time of MOTION then `x=ut` and `y=1/2"gt"^(2)` From equation (i) we get `t=x/u` `:.y=1/2tg(x/u)^(2)=(gx^(2))/(2u^(2))`........iii From figure `y/x=tan theta` or `y=x tan theta`........IV From equation iii and iv we have `x tan theta=(gx^(2))/(2u^(2))` or `x[(gx)/(2u^(2))-tan theta]=0` As x=0 is not possible So `[(gx)/(2u^(2))-tan theta]=0` or `x=(2u^(2)tan theta)/g`...........v Now from equation i, iv and v we get `R=sqrt(x^(2)+y^(2))=sqrt(x^(2)+(x tan theta)^(2))=xsqrt(1+tan^(2)theta)=x SEC theta` or `=(2u^(2))/g "tan" theta sec theta`