Q1: If a particle is projected inside a horizontal tunnel which is 554 cm high with a velocity of 60 m per sec, the angle of projection for maximum range, is

ANS:C - 10°

The answer is required for maximum range, not for maximum height.

For the Maximum range (149.413 m) the angle must be 12 degrees.

Range values for other given angles are less than the range occupied if the angle of trajectory is 12 degrees. This is a projectile motion how can we use this equation. The equation of max height of the projectile is (v^2sin^2angle)/(2g) right? H=u^2 sin^2angle/2g.

5.54=60^2sin^2angle/(2*9.81),
after solving this angle is 10.00650.
So, the answer is 10.