The projectile is fired with a velocity of magnitude 90m/s at angle 45 degree to the horizontal. We can divide the initial velocity into a horizontal component of 90 sin 45 = 90/sqrt 2 and a vertical component of 90 cos 45 = 90/sqrt 2.
The downward acceleration due to gravity of 10 m/s^2 acts on the vertical component and reduces it. The projectile reaches a vertical velocity equal to 0 at the highest point and when it returns to ground level, the magnitude of velocity is 90/sqrt 2 though it is in the opposite direction. We can use this to calculate the time that the projectile was in motion.
- 90/sqrt 2 = 90/sqrt 2 – 10*t
=> 2*90/ sqrt 2 = 10*t
=> t = 2*9/sqrt 2
=> t = 18/sqrt 2
The horizontal distance travelled with a velocity 90/sqrt 2 in 18/sqrt 2 sec is equal to
(90/sqrt 2)*(18/sqrt 2) = 90*18 / 2 = 90*9 = 810 m.
Therefore the projectile travels a horizontal distance of 810 m.