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

Consider the adjacent diagram.

(a) Initial velocity in
x-direction , `u_(x)=u+v_(0) cos theta`
`u_(y)`=Initial velocity in Y - direction
`=v_(0) sin theta`
where angle of projection is `theta`.
Now, we can write
`tan theta =(u_(y))/(u_(x))=(u_(0)sin theta)/(u+u_(0)cos theta)`
`rArr " " theta= tan ^(-1) ((v_(0)sin theta)/(u+v_(0)cos theta))`
(b) Let T be the time of flight.
As net displacement is zero over time period T.
`y = 0, u_(y) = v_(0) sin theta, a_(y)= - g, t = T`
We know that `" " y = u_(y)t+(1)/(2) a_(y)t^(2)`
`rArr " " 0=v_(0)sin theta T + (1)/(2) (-g) T^(2)`
`rArr " " T[v_(0)sin theta-(g)/(2) T] = 0 rArr T= 0, (2v_(0) sin theta)/(g)`
`T=0_(1)`, corresponds to point O.
Hence, `" " T=(2u_(0)sin theta)/(g)`
(c) Horizontal range ,`R=(u+v_(0)cos theta)T=(u+v_(0)cos theta)(2v_(0)sin theta)/(g)`
`=(v_(0))/(g)[2usin theta+v_(0)sin 2 theta]`
(d) For horizontal range to be maximum, `(dR)/(d theta)=0`
`rArr " " (v_(0))/(g)[2ucos theta+v_(0)cos 2 thetaxx2]=0`
`rArr " " 2ucos theta + 2v_(0) [2 cos^(2) theta-1]=0`
`rArr " " 4v_(0)cos^(2)theta+2ucos theta-2v_(0)=0`
`rArr " " 2v_(0)cos^(2)theta+ucostheta-v_(0)=0`
`rArr " " cos theta=(-u+-sqrt(u^(2)+8v_(0)^(2)))/(4v_(0))`
`rArr " " theta_("max")=cos^(-1)[(-u+-sqrt(u^(2)+8v_(0)^(2)))/(4v_(0))]`
`=cos^(-1)[(-u+-sqrt(u^(2)+8v_(0)^(2)))/(4v_(0))]`
(e) If `u=v_(0)`,
`cos theta=(-v_(0)+-sqrt(v_(0)^(2)+8v_(0)^(2)))/(4v_(0))=(-1+3)/(4)=(1)/(2)`
`rArr " " theta=60^(@)`
`"If" u lt lt v_(0), "then " 8 v_(0)^(2)+u^(2)~~8v_(0)^(2)`
`theta_("max")=cos^(-1)[(-u+-2sqrt(2)v_(0))/(4v_(0))]=cos ^(-1)[(1)/(sqrt(2))-(u)/(4v_(0))]`
`"If " u lt lt v_(0), "then "theta_("max")=cos^(-1)((1)/(sqrt(2)))=(pi)/(4)`
`"If" u gt u_(0) " and "u gt gt v_(0)`
`theta_("max")=cos^(-1)[(-u+-u)/(4v_(0))]=0rArr theta_("max")=(pi)/(2)`
(f) `"If" u=0, theta_("max")=cos^(-1)[(0+-sqrt(8v^(2)0))/(4v_(0))]=cos^(-1)((1)/(sqrt(2)))=45^(@)`