1.

Keeping the velocity of projection constant, the angleof projection is increased from 0 to 90°, then thehorizontal range of the projectile30.(1) goes on increasing upto 90°(2) increases upto 45° and decreases afterwards(3) decreases upto 90(4) decreases upto 45° and increases afterwards.

Answer»

the horizontal range depends is proportional to sin2θ , where θ is the projection angle. Thus for θ =45 degrees the horizontal range is maximum as sin90 degrees =1



Discussion

No Comment Found

Related InterviewSolutions