Question

I have a projectile problem with known launch velocity, and target coordinates. How to I determine correct launch angle?
Full Text: If a gunner targets a ship that is 435 meters horizontally from the guns and he knows that the cannon has a muzzle velocity of 68 m/s, at what angle (or angles) should the cannon be aimed? Ignore air resistance, treat g=9.8m/s^2 and that the cannon and ship are at the same height. What angle should the cannon be set to?

Answer 1

what do u know abt projectiles?

Answer 2

v=initial velocity
A=angle from horizontal
t=time
x= distance

For these type of questions, you have to set the angle such that v*sinA is just enough to keep the particle uplifted for t seconds, t being the time taken by a particle of velocity v*cosA to travel the given horizontal distance.

v and x are given.

So you will get 2 equations with 2 unknowns, namely t and A.

1) v*cosA * t = x ------ 1

2) v*sinA - gt = -v*sinA (since a particle thrown up with V velocity, comes back to the starting point with equal and opposite velocity)
or, 2v*sinA = gt ------ 2

Solving 1 and 2, you'll get A.

Answer 3

Thanks for your advice but I'm still having trouble. When trying to solve these equations I have in turn tried isolating x, then t with each equation, then substituting into the other equation. With each attempt I've ended up with a muddled mess!?! Can I ask for more help?

Answer 4

please tell how you approached to solve the ques @KellyThomas ...i will try to correct your mistake..

Answer 5

If I take (1) and isolate t:
v * cosA * t = x
t = x / (v * cosA)

Then substitute into (2):
2v * sinA = gt
2v * sinA = g * (x / (v * cosA))
2v * sinA = gx / (v * cosA)
2v * sinA * v * cosA = gx
sinA * cosA = gx / (2v^2)

At this point all known variables are on the right, and only unknowns are on the left but I don't know how to resolve sinA * cosA = c.

Is there an easier way to approach this? Or do I just need to know how to complete the final step?

Answer 6

well you did great..and the way is also correct ..
now at the end...sinA*cosA=c....
i.e only mathematical part left...and since you have come this far try to do that again... jus focus trigonometry...the basic formulas..and you will get your angle 'A'...
or simply tell me how you tried to do this final step...but i want you to do it yourself so first try..

Answer 7

Remember, 2*sinA*cosA = sin2A.

Use that to simplify your solution.

Answer 8

Thanks guys...
OK, so i did some more research and found that:
sinX * cosY = (sin(X-Y) + sin(X+Y)) / 2
So when X = Y = A:
sinA * cosA = sin(2A) / 2

Substitute back into final equation above:
sinA * cosA = gx / (2v^2)
sin(2A) / 2 = gx / (2v^2)
sin(2A) = 2gx / (2v^2)
sin(2A) = gx / (v^2)
2A = asin( gx / (v^2) )
A = asin( gx / (v^2) ) / 2

So I plugin in the known values:
A = asin( 9.8 * 435 / (68^2) ) / 2
A = 0.5865r

I then submit this answer and I'm told they are expecting an answer containing more than one angle. ...now I'm not sure at which point I should be solving for multiple values.

Answer 9

There will always be more than one value if you're taking the inverse sin of a value. You have to choose the one between 0 and 90 in degrees.

Answer 10

I think they mean that there are two arcs that will impact with the target, one shallow with a velocity closer to the horizontal, the other more vertical. They are taking a single angle as only half the answer.

Answer 11

Oh sorry, yes. I thought you knew. A particle thrown with angle A will land at the same place as a particle thrown with angle 90-A. For same initial velocity.

Answer 12

Fantastic, thanks for your help!