A particle is projected from the surface of earth with velocity equal to its escape velocity at ${45}^{°}$ with horizontal. The angle of its velocity with horizontal at height $h=R,$ where horizontal at some point means a line parallel to tangent on earth just below that point, is ____ (in degree)

At height h = R, the distance of the particle from centre of earth is r = 2R.
Let v be the velocity at that point. Then by law of conservation of mechanical energy, we have 
                                                                                                                          
$\begin{array}{l}{\left(\text{U+K}\right)}_{\text{surface}}\text{=}{\left(\text{U+K}\right)}_{\text{at\hspace{0.17em}\hspace{0.17em}h}}\\ \Rightarrow -\frac{\text{GMm}}{\text{R}}\text{+}\frac{\text{1}}{\text{2}}{\text{mv}}_e^{\text{2}}\text{= -}\frac{\text{GMm}}{\text{R+R}}\text{+}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\end{array}$
Since ,  ${\text{v}}_{\text{e}}\text{=}\sqrt{\frac{\text{2GM}}{\text{R}}}$

$\begin{array}{l}\Rightarrow \frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\text{=-}\frac{\text{GMm}}{\text{R}}\text{+}\frac{\text{GMm}}{\text{R}}\text{+}\frac{\text{GMm}}{\text{2R}}\\ \Rightarrow \text{v = }\sqrt{\frac{\text{GM}}{\text{R}}}\end{array}$
Let $\text{θ}$  be the angle of its velocity with horizontal, then by law of conservation of angular momentum about centre of earth, we have 
  $\begin{array}{l}{\text{mv}}_{\text{e}}\text{Rcos}\left({\text{45}}^{\text{°}}\right)\text{= mvr cos θ}\\ \Rightarrow \text{cosθ\hspace{0.17em}=}\frac{{\text{v}}_{\text{e}}\text{Rcos}\left({\text{45}}^{\text{°}}\right)}{\text{vr}}\\ \Rightarrow \text{cos θ = }\frac{\text{R}\sqrt{\frac{\text{2GM}}{\text{R}}}\frac{\text{1}}{\sqrt{\text{2}}}}{\left(\text{2R}\right)\sqrt{\frac{\text{GM}}{\text{R}}}}\text{=}\frac{\text{1}}{\text{2}}\\ \Rightarrow {\text{θ = 60}}^{\text{°}}\end{array}$