# Nah, Not pi by 4

**Classical Mechanics**Level 4

The angle of projection that gives the maximum horizontal range of a projectile, if the point of projection is at \(h=160 m \) above the point of landing is \(\theta \) for the speed of projection \(u=\sqrt {200} m/s\).

If \(\sin \theta =\frac {\sqrt {a}}{b} \) in simplified form, then find a+b?

Details :

\((a, b) \) are whole numbers.

\(a\) has no factor of a perfect square.

\(g=10 m/s^2\).Neglect any other force.

Horizontal range is the horizontal component of displacement from starting point to landing point.

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