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 = 10m/s^2.)

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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.**

First you need to find the time of flight:

y = v_yo t + (1/2) g t^2

0 = (90 sin 45) t + (5) t^2

factor out a t

0 = t (90 sin45 + 5 t)

set roots equal to zero

t = 0 and 63.6 + 5 t = 0

t = 63.6/5 = 12.7 sec

To find distance use

x = v_xo t

x = 90 cos 45 * 12.7

x = 808 m

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