A projectile is fired at an …

Time of flight: It is the total time for which the projectile remains in air. We know, at the end of time of flight, the projectile reaches the point on the ground. So, vertical height gained by the y = 0. From vertical equation of motion, we have: y=(usinθ)T–12gT2 Now, 0=(usinθ)T–12gT2 ∴T=2usinθg…….(i) Equation (i) gives the time of flight of the projectile for velocity of projection u at an angle θ. Horizontal range: The total distance covered by the body in projectile is called horizontal range. Horizontal range = Horizontal component of velocity × time of flight or,R=ucosθ×2usinθg or,R=u2sin2θg Maximum horizontal range: The horizontal range will be maximum when sin2θ is maximum. i.e.,sin2θor,2θor,θ=1=sin90o=90o=45o Maximum Height: The vertical component of velocity at maximum height is 0. So, V2y0H=(usinθ)2–2gH=u2sinθ−2gH=u2sin2θ2g Two angle of projections for the same range: The range R for velocity of projection u and angle of projection θ is: R=u2sin2θg If (90-θ) be the another angle of projection, then: R2∴R1=R2=u2sin2(90–θ)g=u2sin(180–2θ)g=u2sin2θg