Answer to Question #4301 in Mechanics | Relativity for Charlie
of the arrow to be the origin.
Let& t& be the time of arrow fly, and&
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
& Vv = 75 sin(A).
Notice that the arrow moved horizontally
with constant velocity& Vh& the time t and pass the distance 200 feet.
& 200 = t * 75 cos(A)
& t = 200/(75 cos(A)) = 8/(3
On the other hand, the arrow moves vertically with constant
& g = -9.8 m/s = - 9.8 * 3.2808 ft/s = -32.15 ft/s
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
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)
5 cos(A)^2 + 200 sin(A) cos(A)& = 114.489
& 1 + cos(2A) = 2 cos(A)^2
& 2 sin(A) cos(A) =
2.5 (1 + cos(2A)) + 100 sin(2A)& = 114.489
2.5 cos(2A) + 100 sin(2A)& = 114.489 - 2.5 = 111.989
cos(2A) + 100 sin(2A)& = 111.818
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:
cos(B) = 0.9996876
B = acos(0.9996876) =
0.02499479 rad = 1.432096 degree
let us divide the equation by this
2.5/100.0312 * cos(2A)& +& 100/100.0312 * sin(2A)& =
0.02499 * cos(2A) + 0.9996876 * sin(2A) =
sin(B) * cos(2A) + cos(B)* sin(2A) = 1.119541
hand side is equal to sin(B+2A), so
(*) sin(B+2A) =& 1.119541
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
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