Answer to Question #4301 in Mechanics | Relativity for Charlie

Question

An archer shoots an arrow into the air with an initial velocity of 75 feet per second. The initial height of the shot is 5 feet. The arrow strikes the ground precisely 200 feet from the archer. At what angle did the archer shoot the arrow?

Answer 1

Choose the positive direction of vertical velocity "up", and the starting point
of the arrow to be the origin.

Let& t& be the time of arrow fly, and&
A
be angle at which the archer shoot the arrow.
Then the initial
horisontal velocity of the arrow is
& Vh = 75 cos(A),
and the vertical
one is
& Vv = 75 sin(A).

Notice that the arrow moved horizontally
with constant velocity& Vh& the time t and pass the distance 200 feet.
Then

& 200 = t * 75 cos(A)
whence
& t = 200/(75 cos(A)) = 8/(3
cos(A)).

On the other hand, the arrow moves vertically with constant
acceleration
& g = -9.8 m/s = - 9.8 * 3.2808 ft/s = -32.15 ft/s
and the
initial velocity Vh.
When it strikes the ground, its y-coordinate is -5,
therefore we have the identity:

-5 = Vh t - g t^2 /2
&
-5 =
75 sin(A) * 8/(3 cos(A)) - 32.2& [& 8/(3 cos(A)& ]^2 / 2

-5 = 200
sin(A)/cos(A) - -114.489 / cos(A)^2

-5 cos(A)^2 = 200 sin(A) cos(A)
-114.489
&
5 cos(A)^2 + 200 sin(A) cos(A)& = 114.489


using
the identities
& 1 + cos(2A) = 2 cos(A)^2
& 2 sin(A) cos(A) =
sin(2A)
we get

2.5 (1 + cos(2A)) + 100 sin(2A)& = 114.489


2.5 cos(2A) + 100 sin(2A)& = 114.489 - 2.5 = 111.989

2.5
cos(2A) + 100 sin(2A)& = 111.818

Notice that

sqrt(2.5^2 +
100^2) = 100.0312

Then the numbers
2.5/100.0312 and 100/100.0312

can be regarded as sin and cos of some angle B:
sin(B) =
0.02499
cos(B) = 0.9996876

Hence
B = acos(0.9996876) =
0.02499479 rad = 1.432096 degree
&
let us divide the equation by this
number 100.0312:

2.5/100.0312 * cos(2A)& +& 100/100.0312 * sin(2A)& =
111.989/100.0312
&
0.02499 * cos(2A) + 0.9996876 * sin(2A) =
1.119541

sin(B) * cos(2A) + cos(B)* sin(2A) = 1.119541

The left
hand side is equal to sin(B+2A), so
(*) sin(B+2A) =& 1.119541
which is
impossible.

This means that the assumptions of the problem are NOT
CONSISTENT and the described situation is NOT POSSIBLE.

On the other
hand, if we increase the initial velocity (75) and the height (5) , and decrease
the distance (200) then
the problem would have the solution, and reduces to
solution of (*) with respect to A