### Q.32(b):- Shows that the projection angle θ^{0} for a projectile launched from the origin is given by θ_{0} = tan^{-1}(4h_{m} **/** R)

where the symbols have their usual meaning.

**Answer:- **

Maximum vertical height, h_{m} = u_{0}^{2} Sin^{2}θ / 2g …(i)

Horizontal range, R = u_{0}^{2} Sin^{2}2θ / g … (ii)

h_{m} / R = Sin^{2}θ / 2Sin^{2}2θ

= Sin θ X Sin θ / 2 X 2SinθCosθ

= Sin θ / 4 Cos θ = tan θ / 4

tan θ = (4h_{m} / R)

θ = tan^{-1} (4h_{m} / R)