As, R = \(\frac{v^2_0}{g}\) …(1)
Displacement along y-axis,
−ℎ = ( sin θ) + \(\frac{1}{2}\)(−g)2 …(2)
Displacement along x – axis
= \(\frac{(R+ ∆x)}{v_0cos\theta}\) …(3)
Substituting t in eq. (3),
ℎ = (−v0 sin θ) × \((\frac{R×∆x}{v_0cos\theta})+\frac{1}{2}g(\frac{R×∆x}{v_0cos\theta})^2\)
ℎ = −(R + ∆x)tan θ + \(\frac{1}{2}g(\frac{R+∆x}{v_0cos\theta})^2\)
As θ = 45º
ℎ = −(R + ∆x)×1+\(\frac{1}{2} g\)\(\frac{R+ ∆x}{v_0^2(\frac{1}{2})}\)
ℎ = −(R + ∆x)+ \(\frac{(R + ∆x)^2}{R}\) ∵ R = (\(\frac{v^2_0}{g}\))
= −R−∆x + (R + \(\frac{∆x^2}{R}\) + 2∆x) = ∆x +\(\frac{∆x^2}{R}\)
ℎ = ∆x(1+\(\frac{∆x^2}{R}\) ).