Question

An archer tries to hit a target that is 20m away from him. He can release the arrow at 25m/s. Neglecting air resistance, estimate the angle at which the archer should aim to compensate for the fall of the arrow due to gravity.


Please answer clearly and explain the solution clearly and easy to understand. Thanks!
(guaranteed medal for a good answer)

please give the formulas you used and the given and how and why did you use it.
also the given

But if you can explain better without doing this it is ok. ^_^
thanks!

Answer 1

the equation for the range of projectile neglecting air resistance is: \[R = v^2 \sin (2\theta)/g\]
you have the range to the target, and you know the initial speed, and the g is constant g=9.8 m/s^2, so from the above equation, you have 1 unknown variable you need to find that is \[\theta\] rearranging the equation for angle:
\[\theta = \sin^{-1} (Rg/v^2)/2\]
using this equation \[\theta = 9.13 degrees\]

Answer 2

why is it my teacher has a formula
\[R = -v^2 \sin (2\theta)/g\]

???

Answer 3

there is no negative sign for the range formula, you can check the fomula in our text book, or in wikipedia

Answer 4

so will it give me the same answer if i use the formula with sign?

Answer 5

\[R = (-v^2 \sin \theta)/g\]

Answer 6

Since the velocity vector can be split into component parts that correspond to sin(a) for up , and cos(a) for outward; we need to find some angle(a) that will give us a time span that will carry the arrow the given distance as it loses the fight against gravity.

25 sin(a) is our upward vector

25 cos(a) is our outward vector
.......................................................

25cos(a) t = 20 ; solve for t

t = 20/(25 cos(a))
= 4/(5 cos(a))

lets use this solution for "t" in our equation for the fight against gravity.

\(-4.9\ \cfrac{4}{5cos(a)}^2 + 25sin(a)\ \cfrac{4}{5cos(a)}\)= 0

-3.136 sec^2 + 20 tan = 0 ; sec^2= 1+tan^2

-3.136( 1+tan^2) + 20 tan = 0

-3.136 -3.136 tan^2 + 20 tan = 0; this is now a quadratic, which should be no shock since the path is parabolic to begin with. Lets use the quadratic formula to see if we can get any good results...

-(3.136 tan^2 - 20 tan +3.136)= 0

tan = \(\cfrac{20\pm\sqrt{400-4(3.136)^2}}{2(3.136)}\)

tan = \(\cfrac{20\pm\sqrt{4(100-(3.136)^2)}}{2(3.136)}\)

tan = \(\cfrac{20\pm2\sqrt{100-(3.136)^2}}{2(3.136)}\)

tan = \(\cfrac{10\pm\sqrt{100-(3.136)^2}}{3.136}\)

tan = \(\cfrac{10\pm\sqrt{90.165504}}{3.136}\)

tan = \(\cfrac{10\pm9.495551801}{3.136}\)


tan = 6.216693814
or
tan = .1608572064

to which our angles are the inverse tan of those:

a = 9.138 or a = 80.86 ; cross your fingers :)
................................................

if the angle is abt 9.138

-4.9 t^2 +25sin(9.138) t = 0, when t = abt .81027 secs

and 25cos(9.138) (.81027) = 19.999
................................................

if the angle is abt 80.86

-4.9 t^2 +25sin(80.86) t = 0, when t = abt 5.03726 secs

and 25cos(80.86) (5.03726) = 20.004

so both angles would seem to work for me :)

Answer 7

onaogh is correct