A cricket fielder can throw the cricket ball with a speed vo. If he throws the ball while running with speed u at an angle θ to the horizontal, find
(a) the effective angle to the horizontal at which the ball is projected in air as seen by a spectator.
(b) what will be time of flight?
(c) what is the distance (horizontal range) from the point of projection at which the ball will land?
(d) find θ at which he should throw the ball that would maximise the horizontal range as found in (iii).
(e) how does θ for maximum range change if u >vo , u = vo , u < vo?
(f) how does θ in (v) compare with that for u = 0 (i.e.45o )?
Given:
The initial horizontal velocity of the ball with respect to the fielder = vocosθ
The initial vertical velocity of the ball with respect to the fielder = vosinθ
The final horizontal velocity of the ball with respect to the spectator = vx = u + vocosθ
The final vertical velocity of the ball with respect to the spectator = vy = vosinθ

(a) The effective angle to the horizontal at which the ball is projected in air as seen by a spectator can be found as,
tanθ = ![]()
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(b) Let t be the time of flight.
The displacement in y direction is 0.
![]()
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(c) Let R be the horizontal displacement.
![]()
![]()
(d) For R to be maximum,
should be equal to 0.
![]()
![]()
![]()


(e) When u = vo,


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When, u << vo
We can assume,![]()

u << vo, so we can say that![]()
When, u >> vo
We can assume,![]()

(f) When u = 0,
=45![]()
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