A projectile is fired at an angle of 45 degrees to the horizontal. If its initial velocity is 90 m/s, how far does the projectile travel?(Assume g...

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.