A particle is projected from the top of a tower of height H. Initial velocity of projection is u at an angle ` theta` above the horizontal . Particle hits the ground at a distance R from the bottom of tower. What should be the angle of projection `theta` so that R is maximum ? What is the maximum value of R?

Explanation

Refer to below given figure .

We have selected botton of the tower O as the origin . Horizontal direction is X-axis and vertically upward direction is selected as Y-axis A is the point of projection and B is the point where particle hits the ground . Coordinqates of the initial point a are (0. H) and for the final point (R, 0). components of displacement along X-axis and Y-axis can be written as followings:
`S_(x)=R-0=R,`
`S_(y)=0-H=-H`
Acceleration due to gravity is vertically downwards , hence components can be written as follows :
`a_(x)=0,`
`a_(y)=-g`,
Components of initial velocity are as follows :
`u_(x)=u cos theta`,
`u_(y)=u sin theta`,
X-axis: `S_(x)=u_(x)t+(1)/(2)a_(x)t^(2)`
`rArr =(u cos theta) t rArr t =(R)/(u cos theta)`
Y-axis: `S_(y)=u_(y)t+(1)/(2)a_(y)t^(2)`
`rArr-H=(u sin theta) t+(1)/(2)(-g)t^(2)`
Substituting t from equation (i) , we get the following :
`rArr -H=( u sin theta) ((R)/(u cos theta))-(g)/(2)((R)/(u cos theta))^(2)`
`rArr -H=R tan theta-(gR^(2))/(2u^(2))(1+tan^(2) theta)`
`rArr -2u^(2)H=2u^(2)Rtan theta-gR^(2)(1+tan^(2) theta)`
`rArr -2u^(2)H=2u^(2)R tan theta-gR^(2)-gR^(2)-gR^(2)tan^(2) theta`
`rArr gR^(2) tan^(2) theta-2u^(2) R tan theta+(gR^(2)-2u^(2)H)=0`
Discriminate of above equation must be greater than zero for real value of tan `theta` .
`Delta ge 0`
`rArr 4u^(4)R^(2)-4gR^(2)(gR^(2)-2u^(2)H) ge 0`
`rArr u^(4)-g(gR^(2)-2u^(2)H) ge0`
`rArr u^(4)-g^(2)R^(2)+2gu^(2)H ge 0`
`rArr g^(2)R^(2) le u^(4)+2 "gu"^(2)H`
`rArr R le (u)/(g)sqrt(u^(2)+2gH)`
Hence maximum range can be written as followings:
`rArr R_("max")=(u)/(g)sqrt(u^(2)+2gH)`
If discriminant of a quadratic equation `ax^(2)+bx+c=0` is zero , then equal roots can be written as `x=-b//2a` . If we assume R equal to `R_("max")` described by equation (ii), then we can understand that discriminant of the equation (ii) will be zero , hence `tan theta` for the equation can be written as follows:
`tan theta =(2Ru^(2))/(2gR^(2))=(u^(2))/(gR)`
Substituting value of R from equation (iii) , we get the following :
`tan theta=(u^(2))/(g(u)/(g)sqrt(u^(2)+2gH)`
`rArr tan theta=(u)/(sqrt(u^(2)+2gH)`
`rArr theta=tan^(-1)((u)/(sqrtu^(2)+2gH))`