# Nah, Not pi by 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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