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A cricket fielder can throw the Cricket ball with a speed v0. 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 maximize the horizontal range as found in (c).

(e) how does θ for maximum range change if u > v0, u= v0, u < v0?

(f) how does θ in (e) compare with that for u = 0 (i.e., 45°)?

1 Answer

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(a) The angle of projection with horizontal seen by spectator will be

tan θ = \(\frac{u_y}{u_x}=\frac{v_0sin\,\theta}{u+v_0\,cos\,\theta}\)

(b) y = uxt + \(\frac{1}{2}a_yt^2\) , as y = 0,

T = \(\frac{2v_0 sin\,\theta}{g},T[v_0sin\,\theta-\frac{g}{2}T]\) = 0

(c) R = ( u+ v0 cos θ)T = ( u+ v0 cos θ) \(\frac{2v_0 sin\,\theta}{g}\)

(d) \(\frac{dR}{dv}=0\)

\(\frac{v_0}{g}[2ucos\,\theta+v_0cos\,2\theta\times2]\)=0,

cos θ = \(\frac{-u\pm \sqrt{u^2+8v^2_0}}{4v_0}\)

⇒ θ = \(tan^{-1}(\frac{v_0}{u})\)

(e) If u = v0

cos θ = \(\frac{-v_0\pm \sqrt{v^2_0+8v^2_0}}{4v_0}=\frac{-1+3}{4}=\frac{1}{2}\)

⇒ θ = 60º

If u < < v0, then \(8v^2_0+u^2\)\(8v^2_0\)

\(\theta_{max}\) = \(cos^{-1}[\frac{-u\pm2\sqrt2 v_0}{4v_0}]\)

\(cos^{-1}[\frac{1}{\sqrt2}-\frac{u}{4v_0}]\)

If u < < v0 so neglecting \(\frac{u}{4v_0}\) , then

\(\theta_{max}\) = \(cos^{-1}(\frac{1}{\sqrt2})\)

= 45º

If u > v0 and u > > v0

\(\theta_{max}\) = \(cos^{-1}[\frac{v_0}{u}]\) = 0     ∵ [\(\frac{v_0}{u}\) → 0]

\(\theta_{max}\) = 90º

(f) If u= 0, \(\theta_{max}\)\(cos^{-1}[\frac{0\pm \sqrt{8v^2_0}}{4v_0}]\)

= \(\cos^{-1}(\frac{1}{\sqrt2})\) = 45º

Hence, when u = 0, θ ≥ 45º

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