# please explain why this trig solution is wrong

Discussion in 'General Math' started by kjm, Nov 15, 2003.

1. ### kjmGuest

Hi all. I don't understand why my approach to this problem is flawed. Can
someone explain it. Here is the problem and my attempted solution. (I know
how to get the right answer but I don't know why a particular approach does
not work)

From Schaum's Beginning Physics I, Alvin Halpern
Problem 3.43 A marksman is trying to hit a stationary target that is 100 m
above the level of his rifle. He aims the rifle 5 deg above the horizontal,
fires, and hits the target just as the bullet reaches the highest point in
its trajectory. What is the muzzle velocity of the rifle? (not concerned)
How far away, horizontally, was the target?

The conondrum is in my particular solution to the 2nd of the two questions,
how far away? I figured I could determine how far the target was by drawing
a right triangle and applying trig like so:

/| 100m
/ |
/__|
5 deg

tan 5 deg = 100 / x so x = 100 / tan 5 = 1143 m

the correct answer is 2286 meters which is arrived at my applying well known
physics formulae. The correct answer is 2286 m. This is given in the book
and I confirmed it once using I forget what formulas.

Illumination appreciated.

kjm
newsbox AT bwknight.net
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kjm, Nov 15, 2003

2. ### HenryGuest

The clue is in "highest point in its trajectory" suggesting that you
need to take account of gravity

Henry, Nov 15, 2003

3. ### Tom LorenzoGuest

You have to understand more physics to get what the problem is about.
Meaning you need to be less dependent on just finding a formula to apply and
put some effort into that understanding. As Henry noted, the key is noting
that the bullet strikes the target at the height of its trajectory. For
these kinds of problems, you ignore air resistance to keep the calculations
simple.

Think of it this way -- at the height of its trajectory, the vertical
component of velocity of the bullet is zero. From that point of zero
vertical velocity, the only thing affecting the vertical velocity of the
bullet is gravity. So how long would it take a bullet dropped from 100
meters to hit the ground? A further understanding of the problem is to
realize that for level ground (no air resistance), the bullet will take as
long to get from the ground to the height of its trajectory as it takes to
fall from its highest point to the ground. By the way, your comment that you
were not concerned about the muzzle velocity is a good indication you do not
have a good grasp of the physics. Because after you compute how long it
would take to fall 100 meters, the next step is to compute that muzzle
velocity based on generating an equation for the vertical velocity of the
bullet taking into account the "upward" velocity from the rifle and the
"downward" velocity from the acceleration of gravity. Knowing the time it
takes to fall 100 meters allows you to solve for the component of muzzle
velocity in the vertical direction, and then the muzzle velocity, and then
the component of muzzle velocity in the horizontal direction, and finally
computing how far it traveled in the horizontal direction (because at that
stage you know both the horizontal velocity and how much time it traveled at
that velocity).

Regards
- Sox

Tom Lorenzo, Nov 16, 2003
4. ### kjmGuest

I have received an email that explains how my approach is flawed. I did not
take into account that the bullet travels along a parabolic trajectory when
drawing my diagram. I am satisfied with this explanation but welcome more
input if anyone wishes to say more.

kjm
newsbox AT bwknight.net
replace " AT " with @

....cut
....cut

kjm, Nov 17, 2003