General

Pretty much what LowWillPower said:
Given: vi = 1200 m/s and d=500m; vx=1200cos(w) and vy=1200sin(w)
1. Use d = v*t for x-direction because we know {vx=1200cos(w)} doesn't change and d is also given (d=5000). And solve for t. We know t is same for either x or y-comp.
__ t = 5000 / {1200cos(w)}
2. We also know change in y=0 because it started and landed at ground level: delta y = vy*t + 1/2 a*t^2 becomes:
__ -vy*t = (1/2)*at^2 (a= -9.81m/s^2 so it will cancel out the negative signs). Plug in t and solve for w:
__1200sin(w) * [5000 / {1200cos(w)}] = (1/2)*(9.81)*[5000 / {1200cos(w)}]^2, simplifying we get:
__2sin(w)cos(w) = 9.81*5000 / (1200^2)
__[b]sin(2w) = 9.81*5000/(1200^2)[/b]. Solve for w.
__[b]*note: there should be 2 values for w (your angle), for every projectile motion angle, there is another complementary angle that gives the same results.[/b]

42 degrees, if you know what I mean.

http://puu[.]sh/25ab6

[quote=LowWillpower]Projectile motion. I'm guessing we assume no air drag. Solve for the x and y directions. You know in x there is no acceleration, so distance x = x_velocity*time

We know the magnitude of velocity, but the x_velocity will be velocity*cos(angle), while y_velocity is velocity*sin(angle).

We also know in the y direction the projectile will have gravity and velocity acting on it. distance y = (g* (t^2))/2 + y_velocity*time We assume no initial vertical displacement, since that is an arbitrary axis. Also, I'm guessing we assume the 5000m away target is at the same height as us, in which case we want when the distance in the y direction is 0. plug 0 in to the y equation so 0 = (g*(t^2))/2 +y_velocity*time. The obvious answer is t = 0, which is when we start at 0 height, but if you rearrange (by removing t from both terms) you can get t = -y_velocity*2/g. Remember g is negative. Plug t into the other equation. You get x = x_velocity*y_velocity/4.9

We already know x, now we just need x_velocity or y_velocity, which we can find from the velocity and angle relationships. 4.9x = vcos(angle)*vsin(angle)

Rearrange and replace x and v 4.9*5000/1200= cos(angle)sin(angle)

Now we know sin(a+b) = sin(a)cos(b)+sin(b)cos(a), but if a = b then sin(2a) = 2sin(a)cos(a), so sin(2a)/2= sin(a)cos(a), substituting into what we have (and multiplying the 2 on the bottom over)

9.8*5000/1200 = sin(2*angle), take the inverse 2*angle = sin^(-1)(9.8*5000/1200)

Divide by 2, find the angle.[/quote]

Ohhh I get it now. I forgot about trig identities lol. Thank you

[url=http://www.youtube.com/watch?v=6gLMSf4afzo]Ain't Nobody Got Time For That[/url]

[quote=LowWillpower]Projectile motion. I'm guessing we assume no air drag. Solve for the x and y directions. You know in x there is no acceleration, so distance x = x_velocity*time

We know the magnitude of velocity, but the x_velocity will be velocity*cos(angle), while y_velocity is velocity*sin(angle).

We also know in the y direction the projectile will have gravity and velocity acting on it. distance y = (g* (t^2))/2 + y_velocity*time We assume no initial vertical displacement, since that is an arbitrary axis. Also, I'm guessing we assume the 5000m away target is at the same height as us, in which case we want when the distance in the y direction is 0. plug 0 in to the y equation so 0 = (g*(t^2))/2 +y_velocity*time. The obvious answer is t = 0, which is when we start at 0 height, but if you rearrange (by removing t from both terms) you can get t = -y_velocity*2/g. Remember g is negative. Plug t into the other equation. You get x = x_velocity*y_velocity/4.9

We already know x, now we just need x_velocity or y_velocity, which we can find from the velocity and angle relationships. 4.9x = vcos(angle)*vsin(angle)

Rearrange and replace x and v 4.9*5000/1200= cos(angle)sin(angle)

Now we know sin(a+b) = sin(a)cos(b)+sin(b)cos(a), but if a = b then sin(2a) = 2sin(a)cos(a), so sin(2a)/2= sin(a)cos(a), substituting into what we have (and multiplying the 2 on the bottom over)

9.8*5000/1200 = sin(2*angle), take the inverse 2*angle = sin^(-1)(9.8*5000/1200)

Divide by 2, find the angle.[/quote]

I think you are wrong. because (9.8*5000/1200) = 40.83 and Sin^-1(40.83)= impossible because ArcSin only works for a domain of [-1,1]

[quote=LowWillpower]Projectile motion. I'm guessing we assume no air drag. Solve for the x and y directions. You know in x there is no acceleration, so distance x = x_velocity*time

We know the magnitude of velocity, but the x_velocity will be velocity*cos(angle), while y_velocity is velocity*sin(angle).

We also know in the y direction the projectile will have gravity and velocity acting on it. distance y = (g* (t^2))/2 + y_velocity*time We assume no initial vertical displacement, since that is an arbitrary axis. Also, I'm guessing we assume the 5000m away target is at the same height as us, in which case we want when the distance in the y direction is 0. plug 0 in to the y equation so 0 = (g*(t^2))/2 +y_velocity*time. The obvious answer is t = 0, which is when we start at 0 height, but if you rearrange (by removing t from both terms) you can get t = -y_velocity*2/g. Remember g is negative. Plug t into the other equation. You get x = x_velocity*y_velocity/4.9

We already know x, now we just need x_velocity or y_velocity, which we can find from the velocity and angle relationships. 4.9x = vcos(angle)*vsin(angle)

Rearrange and replace x and v 4.9*5000/1200= cos(angle)sin(angle)

Now we know sin(a+b) = sin(a)cos(b)+sin(b)cos(a), but if a = b then sin(2a) = 2sin(a)cos(a), so sin(2a)/2= sin(a)cos(a), substituting into what we have (and multiplying the 2 on the bottom over)

9.8*5000/1200 = sin(2*angle), take the inverse 2*angle = sin^(-1)(9.8*5000/1200)

Divide by 2, find the angle.[/quote]

ah physics music to my ears.

Projectile motion. I'm guessing we assume no air drag. Solve for the x and y directions. You know in x there is no acceleration, so distance x = x_velocity*time

We know the magnitude of velocity, but the x_velocity will be velocity*cos(angle), while y_velocity is velocity*sin(angle).

We also know in the y direction the projectile will have gravity and velocity acting on it. distance y = (g* (t^2))/2 + y_velocity*time We assume no initial vertical displacement, since that is an arbitrary axis. Also, I'm guessing we assume the 5000m away target is at the same height as us, in which case we want when the distance in the y direction is 0. plug 0 in to the y equation so 0 = (g*(t^2))/2 +y_velocity*time. The obvious answer is t = 0, which is when we start at 0 height, but if you rearrange (by removing t from both terms) you can get t = -y_velocity*2/g. Remember g is negative. Plug t into the other equation. You get x = x_velocity*y_velocity/4.9

We already know x, now we just need x_velocity or y_velocity, which we can find from the velocity and angle relationships. 4.9x = vcos(angle)*vsin(angle)

Rearrange and replace x and v 4.9*5000/1200= cos(angle)sin(angle)

Now we know sin(a+b) = sin(a)cos(b)+sin(b)cos(a), but if a = b then sin(2a) = 2sin(a)cos(a), so sin(2a)/2= sin(a)cos(a), substituting into what we have (and multiplying the 2 on the bottom over)

9.8*5000/1200 = sin(2*angle), take the inverse 2*angle = sin^(-1)(9.8*5000/1200)

Divide by 2, find the angle.

Use a scope.