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=160mh=160 m above the point of landing is θ\theta for the speed of projection u=200m/su=\sqrt {200} m/s.

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

Details :

  • (a,b)(a, b) are whole numbers.

  • aa has no factor of a perfect square.

  • g=10m/s2g=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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