Parabolic Trajectory Problems

This topic is best explained by way of example. Consider the situation shown in this diagram: a projectile is launched from the ground toward a point on a cliff at relative height 400 m, at range 300 m measured horizontally. The cannon has a "muzzle velocity" 600 km/h. There are two questions we will address (ignoring air friction):

  1. If the cannon were aimed straight at the target, where would the projectile actually land? (path shown in red)
  2. To make the projectile hit the target, at what projection angle should the cannon be aimed? (path shown in green)

If you saw the movie Mulan, you know that there was little time for aiming of field artillery in the mountain-pass battle scene (fortunately we have computerized equipment today). When a projectile moves near the earth's surface, the motion along the horizontal (x) is constant velocity while the motion along the vertical (y) is constant acceleration. The equations of interest are:

x = (vo)xt       and       y = (vo)yt - 0.5 gt2.

First, we have to convert the muzzle velocity to m/s: (600)(0.278) = 167 m/s. Then, we calculate q1 from tan(q1)=400m/300m; q1=53o.

Next, we look at the two equations of motion to see what we can solve for. The time of flight t can be obtained first. Using (vo)x=167cos53o =101m/s, we get t=(300/101)=2.97s.

Finally, we plug this value of t into the y equation. Using (vo)y=167sin53o =133m/s, we get
y=(133)(2.97)-(4.9)(2.97)2=352 m. So the answer to the first question is: aiming right at the target causes us to miss it (we hit a point 48 m below it).

To answer the second question, we could use the x equation to solve for t:

x=300=167(cosq2)t; t=1.80/(cosq2).

Then the y equation becomes:
y=400=167(sinq2)t-4.9t2.

Substituting the previous expression for t gives:
400=300tanq2-(15.9/cos2q2).

Solving this for q2 is too difficult (it would require using trigonometric identities and/or series expansions). As an alternative, let's aim for a point 48 m above the target. Using trigonometry, tanq2=448/300 which gives q2=56o. Now we can use the question #1 equations over again; when this is done we get y=396m (very close to the desired value of 400m).

It would be interesting to figure out if the projectile is still rising or beginning to fall when it reaches the target. Using q2=56o, let's determine the highest altitude of the projectile and the time of flight in getting there.

The first equation of interest is (v)y=(vo)y-gt. At the top of the motion, this is equal to zero. So 0=167(sin56o)-9.8t, which gives us t=14.1s!

Then substituting into y=(vo)yt-0.5gt2 gives y=978m.

Clearly, the projectile is on its way up when it runs into the cliff and the green path in the drawing is not accurate.

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