(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