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March 30, 2015

March 30, 2015

Posted by **hanna tret** on Friday, September 9, 2011 at 11:31am.

- trig -
**tchrwill**, Saturday, September 10, 2011 at 9:49amProjectile Motion

The ballistic analysis of a projectile may be simplified by combining the components of uniform and accelerated motion of the projectile. Consider a projectile fired from a gun with a muzzle velocity of V in a direction µº to the horizontal. We now ask ourselves, "How far from the gun will the projectile hit the ground?" and "How high will the projectile go?"

Breaking the velocity up into its horizontal and vertical components makes one thing immediately obvious. Only the vertical component is subject to the force of gravity. Applying the expression for accelerated motion, we can find the time of flight and the maximum height reached. The vertical component of the launch velocity is Vv = Vsin(µ). Therefore, from Vf = Vo - gt, the time to reach maximum height derives from 0 = Vsin(µ) - gt1 or Vsin(µ) = gt1 making

....... t1 = Vsin(µ)/g.

The height reached then derives from h = Vsin(µ)t1 - gt1^2/2 which can be written as h = gt1^2 - gt1^2/2 or

.......h = gt1^2/2.

The time for the projectile to return to the ground derives from

.......-h = -gt2^2/2. Since the height up equals the height down, it becomes clear that t1 = t2 making the total time of flight t = 2t1.

The horizontal distance traveled derives from d = Vcos(µ)t = 2Vt1cos(µ) = 2V[Vsin(µ)/g]cos(µ) making

.......d = V^2sin(2µ)/g.

Now, eliminating t1 from Vsin(µ) = gt1 and h = gt1^2/2, the maximum height reached becomes

.......h = V^2sin^2(µ)/2g.

As with uniform accelerated motion, these expressions ignore atmospheric drag.

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