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 component 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
.
Now solve the equations you now have simultaneously (I did it by substituting
T from vertical motion into horizontal) and I got: 20000=2(1604)2 sinqcosq/g You may know 2sinqcosq º sin(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 .]
