Question

a body is projected an angle theta with respect to horizontal direction with velocity U the maximum range of the body is . Plz friends answer it fast

Answer 1

The range of a projectile is given as ,

Range = u^2 sin 2*theta / g
Well this can be easily derived from the equations of motion from the chapter of kinematics by just breaking up the quantities in their components separately in the x and y axes

Well now the only variable thing in the equation is theta

And well sin 2theta has it's maximum value as 1 which is when we keep the angle as 45° so that then the angle becomes 90°

Thus, the maximum range will be u^2/g.

Answer 2

a projectile moves with speed u and inclined with horizontal is ∅

then, initial velocity vector (U)

= ucos∅i + usin∅ j

let at time = t velocity of projectile is perpendicular upon initial velocity .

then, that time velocity ( V)= ucos∅ i + ( usin∅ -gt) j

now,

U and V are perpendicular so,

dot product of U and V = 0

( ucos∅ i + usin∅ j ){ucos∅i +( usin∅-gt)j } = 0

u² cos²∅ + u²sin²∅ -usin∅gt = 0

u²( cos²∅ + sin²∅ ) = usin∅gt

u = sin∅ gt

t = u/gsin∅

now , put this in V

V = ucos∅i + ( usin∅ -gu/gsin∅)j

= u cos∅i + u( sin²∅ -1)/sin∅ j

= ucos∅i - ucos²∅/sin∅ j

magnitude of V

V = u √( cos²∅ + cos²∅cot²∅)

u√( cos²∅(1 + cot²∅))

=u√cos²∅(cosec²∅)

=ucot∅

hence , V = ucot∅