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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°)? |
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Answer» (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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