PhysicsHard53×since 2003Q7789A projectile is launched at an angle 'α\alphaα' with the horizontal with a velocity 20 ms^−-−1. After 10 s, its inclination with horizontal is 'β\betaβ'. The value of tanβ\betaβ will be : (g = 10 ms^−-−2).Atanα\alphaα + 5secα\alphaαBtanα\alphaα −-− 5secα\alphaαC2tanα\alphaα −-− 5secα\alphaαD2tanα\alphaα +++ 5secα\alphaαCheck answerSkip