The horizontal range of a projectile fired at an angle of 150 is 50 m. If it is fired with the same speed atan angle of 45°, its range will be

Given,  θ = 15o and R = 50mRange, R=u2sin2θgPutting all the given values in the formula, we getR=50 m=u2sin2×15°9⇒50×g=u2sin30°=u2×12⇒50×g×2=u2⇒  u2=50×9.8×2=980⇒ u=980  m/s =145 m/s  For θ=45°,R=u2sin2×45°g=u2g      ∵sin90°=1⇒ R'=(145)29=14×14×59.8=100 m