Hi I am new here so here is my question.
A gun fires a projectile at 2200 ft/s muzzle velocity at a target 2.0 km away, at the same altitude; neglecting wind, what angle above the horizon must you aim, to hit the target? acceleration due to gravity is 9.8 m/s^2.
Ty for any help.
Also where do I sign up? Sorry, but I am having trouble seeing where things are right now (posting this from a voice recognition engine).
3rd question. I am confused about this Y intercept stuff. Can someone please explain it to me?
(if you can't answer the last one, it's ok I really just need an answer for the first question). Ty so much if anyone can help ☺.
+33658 For the horizontal component of motion the velocity is constant, so if the initial angle is θ, we have, using distance = velocity* time, and that 1ft/s = 0.3048m/s, that:
2000 = 670.56*cos(θ)*t ...(1) where t is the flight time.
For the vertical motion we have vertical distance = initial velocity*time + (1/2)*accn*time2
Now the net vertical distance here is zero, so:
0 = 670.56*sin(θ)*t - (1/2)*9.8*t2 ...(2)
Rearrange equation (1) to get t = 2000/(670.56*cos(θ)) and substitute this into (2):
0 = 670.56*sin(θ)*2000/(670.56*cos(θ)) - (1/2)*9.8*(2000/670.56*cos(θ))2
Multiply this through by cos(θ)2 and divide through by 2000 to get:
0 = cos(θ)*sin(θ) - (1/2)*9.8*2000/670.562
or 0 = (1/2)sin(2θ) - (1/2)*9.8*2000/670.562
Multiply through by 2 and add 9.8*2000/670.562 to both sides:
sin(2θ) = 9.8*2000/670.562
so 2θ = sin-1(9.8*2000/670.562)
or θ = sin-1(9.8*2000/670.562)/2
sin360∘−1(9.8×2000670.562)2=1.2491398003115
θ ≈ 1.25°. A very small angle!
To sign up go to the Home page: http://web2.0calc.com/ and next to the Login button at the top right of the page you will be able to select Register.
.
+33658 For the horizontal component of motion the velocity is constant, so if the initial angle is θ, we have, using distance = velocity* time, and that 1ft/s = 0.3048m/s, that:
2000 = 670.56*cos(θ)*t ...(1) where t is the flight time.
For the vertical motion we have vertical distance = initial velocity*time + (1/2)*accn*time2
Now the net vertical distance here is zero, so:
0 = 670.56*sin(θ)*t - (1/2)*9.8*t2 ...(2)
Rearrange equation (1) to get t = 2000/(670.56*cos(θ)) and substitute this into (2):
0 = 670.56*sin(θ)*2000/(670.56*cos(θ)) - (1/2)*9.8*(2000/670.56*cos(θ))2
Multiply this through by cos(θ)2 and divide through by 2000 to get:
0 = cos(θ)*sin(θ) - (1/2)*9.8*2000/670.562
or 0 = (1/2)sin(2θ) - (1/2)*9.8*2000/670.562
Multiply through by 2 and add 9.8*2000/670.562 to both sides:
sin(2θ) = 9.8*2000/670.562
so 2θ = sin-1(9.8*2000/670.562)
or θ = sin-1(9.8*2000/670.562)/2
sin360∘−1(9.8×2000670.562)2=1.2491398003115
θ ≈ 1.25°. A very small angle!
To sign up go to the Home page: http://web2.0calc.com/ and next to the Login button at the top right of the page you will be able to select Register.
.
