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Projectile Angles
By Mrs. Toni Beardon on Tuesday, September 18, 2001 - 08:42 am:

From Jason Trang:
Can someone please answer this question?

A canon ball is fired with an initial speed of 1604m/s at a certain angle.
Once it lands, it has travelled 20km horizontally.
Ignoring wind, air resistance and using the value of 9.807m/s2 for gravity, how can I obtain the angle that the canon ball is shot at?

THANKS

By Jim Oldfield on Wednesday, September 19, 2001 - 12:23 am:

As usual for projectiles, take the vertical componant of motion and write down what you know (u=1604sinq, s=0, t=T, a=-g) then put this into one of the projectile formulae. Next work horizontally (u=1604cosq, s=20000. t=T, a=0) and put these also into a formula. Note I have made up the variables T and q here, whose meanings I hope are clear. The formula I used in both directions was s=ut+½at2.

Now solve the equations you now have simultaneously (I did it by substituting T from vertical motion into horizontal) and I got: 20000=2(16042)sinqcosq/g You may know 2sinqcosqºsin(2q), so you should now be able to solve the equation for q

I believe I'm right in saying that the optimum angle for any projectile to be shot at (ignoring air resistance) is 45°.

Jim


[Editor's Comment: Jim's final comment is indeed correct and can be seen from his own working. By resolving vertical and horizontal components of motion (as Jim has done) obtain the distance travelled (d) in terms of the fixed speed at which the projectile is shot at (v) and the projection angle (a). Now find the particular a which maximises d (by differentiating). Further NRICH projectile questions were discussed at Projectile Questions.]
Projectile