A cricket fielder can throw the cricket ball with a speed ` v_0`. If he throws the ball while running with speed (u) at angle ` theta` to the horizontal, find
(b) what will be time of flight ?
(c ) what is the distance (horizontal range) form the point of projection at which the ball will land ?
(d) find ` theta` at which he should throw the ball that would maxmise the horizontal range range as found in (c ).
(e) how does ` theta` for maximum range change if ` u gt v_0, u=v_0, ult v_0` ?
(f) how does ` theta` in (e) compare with that for ` u=0 (i.e., 45^@) ?

Explanation

Refer Fig, 2 (EP) . 33 ,
(a) Here, at ` O, u_x = u + v_0 cos theta` ltbRgt ` u_y = v_0 sin theta `
` tan theta = u_y /u_x = (v_0 sin theta )/(u + v_0 cos theta)`
or ` theta = tan ^(-1) ( ( v_0 sin theta )/(u + v_0 cos theta))`
.
(b) Let (T) bne the time of flight of cricket ball from (O) to (B), then taking vertical moton of ball form (O) to (B), we have ` y_0 , = 0, u_y = v 0 sin theta , a_y =- g, t= T`
As, ` y= y_0 + u_y t + 1/2a_y t^2`
:. ` 0=0 + v_0 sin theta + 1/2 (-g ) T^2` or ` T= (2 v_0 sin theta ) /g`
` (C) Horizontal range , ` R= ( u + v_0 cos theta ) T = ( u + v_0 cos theta ) ( 2 v_0 sin theta)/g = v_0 /g [ 2 us sin theta + v_-0 sin 2 theta ]`
(d) Horizontal range will be maximu if (dT)/(d theta) =0`
:. ` (d R) /(d theta )_ = v_0 /g [ 2 u cos theta + v_0 cos 2 theta xx 2 ] =0`
or ` 2 u cos theta + 2 v_0 [ 2 cos^2 theta - 1 ] =0` or ` 4 v_0 cos^2 theta + 2 u cos theta - 2 v_0 =0` ltbr. or ` ocs = (- 2 u +- sqrt ( 4 U^2 + 32 b_0^2 )/( 8 v_0) = ( -u +- sqrt ( u^2 + 8 v-0^2 ) /( 4 v_0)` or ` theta _(max = cos^(-1) [ (- u +- sqrt u^2 + 8 v_0^2)/( 4 v_0)]`
(e ) ` (i) I(d ` u = v_0, ten cos theta = ( -v _0 +- sqrt ( v_0^2 + 8 v_0^2) /( 4 v_0) = (- 1 + 3)/4 = 1/2 cos 60^@ ` or ` theta = 60^@`
(ii) If ``v_0`, then ` 8 v_0^2 + u^2 ~~ 8 v_0^2`
` theta _( max = cos^(-1) [ ( -u +- 2 sqrt 2 v_0)/( 4 v_0)] = cos ^(-1) [ 1//(sqrt 2) - u/(4 v_0)]`
If ` u ltlt v_0 then ` theta _(max) = cos ^(-1) [ 1/(sqrt 2) ] = (pi)/4`
(ii) If ` u gt v ` and ` u gtgt v_0`
` theta _max) = cos^(-1) [( 0 +- sqrt 8 v_0^2) /(4 v_0)] = cos^(-1)` (1/(sqrt 2)) = 45^@` ltbr. (f) ` If ` u = 0, theta _(max) = cos ^(-1) [ (0+- sqrt 8 v_0^2)/( 4 v_90))] = cos ^(-1) (1/(sqrt 2) = 45 ^@`.