Given these two formulas solve for theta
0=24+400*sin(theta)*T+(1/2)*(-9.81)*((…
and
12874.752=400*cos(theta)*T
Please show how to arrive at answer, (I know what the answer is (25.9236685 degrees)... I have CAS software, I just want to know how to solve it) when I solve it by hand I keep getting 23.1307 degrees)
10 points for detailed answer
Ps the original problem was to find the initial angle of a projectile launched at 400 m/s from the deck of a ship (24 m) and it travels 8 miles before splashing into the water- no air resistance or friction factored into this
0=24+400*sin(theta)*T+(1/2)*(-9.81)*((…
and
12874.752=400*cos(theta)*T
Please show how to arrive at answer, (I know what the answer is (25.9236685 degrees)... I have CAS software, I just want to know how to solve it) when I solve it by hand I keep getting 23.1307 degrees)
10 points for detailed answer
Ps the original problem was to find the initial angle of a projectile launched at 400 m/s from the deck of a ship (24 m) and it travels 8 miles before splashing into the water- no air resistance or friction factored into this
-
From your description of the original problem - the angle calculated by ignoring the added fall of 24 m to the water would be VERY close to actual answer. The reason is the large RANGE of this projectile 8 mi = 12,875 m. The final fall of 24 m is going to take such a small time as to almost negligibly affect the initial angle = Θ.
so just to get U close:
Let unknown angle = Θ
Vo = initial velocity = 400 m/s
Voy = initial VERTICAL velocity = Vo sin Θ = (400)(sin Θ)
Vox = initial HORIZONTAL velocity = CONSTANT = (400)(cos Θ)
t = time to max height above release point (deck) = Voy/g = 400/g(sin Θ)=40.7747sinΘ
2t = flight time ignore fall from deck to water = 81.5494sinΘ
R = range of projectile = 8 mi = 12,875 m {google conversion}
12,875/Vox = 12,875/(400)(cosΘ) = 2t = 32.1875/(cosΘ)
2t = 81,5494sinΘ
32.1875/(cosΘ) = 81.5494(sinΘ)
sinΘcosΘ = 0.3947
2sinΘcosΘ = sin2Θ = (2)(0.3947) = 0.7894
arcsin2Θ = 0.7894
2Θ = 52.129
Θ = 26.06° <= close but of course not the actual angle
However by trigonometry U could now estimate amount of RANGE change due to THIS angle as being = 24(cos 26.06°) = 21.6 m. So the amount of distance change from the RANGE would be:
12,875 - 22 = 12,853 and THIS would be the symmetric end of the projectile's actual path to the level of the deck - Recalculate the above using THIS as the RANGE number and the angle Θ will be the correct one.
This is not what 'math folks' call an "elegant solution" but it does solve :>)
so just to get U close:
Let unknown angle = Θ
Vo = initial velocity = 400 m/s
Voy = initial VERTICAL velocity = Vo sin Θ = (400)(sin Θ)
Vox = initial HORIZONTAL velocity = CONSTANT = (400)(cos Θ)
t = time to max height above release point (deck) = Voy/g = 400/g(sin Θ)=40.7747sinΘ
2t = flight time ignore fall from deck to water = 81.5494sinΘ
R = range of projectile = 8 mi = 12,875 m {google conversion}
12,875/Vox = 12,875/(400)(cosΘ) = 2t = 32.1875/(cosΘ)
2t = 81,5494sinΘ
32.1875/(cosΘ) = 81.5494(sinΘ)
sinΘcosΘ = 0.3947
2sinΘcosΘ = sin2Θ = (2)(0.3947) = 0.7894
arcsin2Θ = 0.7894
2Θ = 52.129
Θ = 26.06° <= close but of course not the actual angle
However by trigonometry U could now estimate amount of RANGE change due to THIS angle as being = 24(cos 26.06°) = 21.6 m. So the amount of distance change from the RANGE would be:
12,875 - 22 = 12,853 and THIS would be the symmetric end of the projectile's actual path to the level of the deck - Recalculate the above using THIS as the RANGE number and the angle Θ will be the correct one.
This is not what 'math folks' call an "elegant solution" but it does solve :>)