a projectile is fired vertically at an angle theta from ground with velocity v. speed of projectile when it is moving at an angle delta from horizontal is ​  

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when projectile fired vertically with an angle θ from the ground it's horizontal and vertical component of velocity are given as  vx=vcosθ and vy=vsinθ. Also there is no acceleration in the horizontal so the horizontal component of velocity does not change during the projectile motion. 
Let after a time t the projectile makes angle δ from the horizontal and let the velocity is u then the horizontal and vertical component of velocity at this time is given by 
since the acceleration in the vertical direction is g so the vertical velocity is given by 
But since u makes the angle δ from the horizontal so 
ux=ucosδuy=usinδ       ortanδ=uyux=vsinθ-gtvcosθvcosθtanδ=vsinθ-gtgt=vsinθ-vcosθtanδso u2=ux2+uy2u2=vcosθ2+vsinθ-gt2u2=vcosθ2+vsinθ-vsinθ+vcosθtanδu2=vcosθ2+vcosθtanδ2u2=v2cos2θ+v2cos2θtan2δu2=v2cos2θ1+tan2δu2=v2cos2θ·sec2δu=vcosθ·secδ

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