# Answer on Mechanics | Relativity Question for Charlie

Question #4301

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?

Expert's answer

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

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

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