Two Dimensional Motion (II)

1. Feb 15, 2007

Aggie

A gun shoots a bullet with a velocity of magnitude 409 m/s. The goal is to hit the target 1,290 meters away. How high above the target must you aim to correct for gravity? (Assume the gun and target are at the same height)

I have attempted solving it but I am stuck at the point below:

D*Tan A = (g/2)(D/Vo)^2(Cos A)^2

where,
D=1290 m
Vo= 409 m/s

I think Tan A is the solution but i do not know how to get to that point

2. Feb 15, 2007

Integral

Staff Emeritus
How is your Trig?

Do you know the definition of TanA in terms of SinA and CosA? Use that to isolate A.

3. Feb 15, 2007

Aggie

That's as far as I can go

4. Feb 15, 2007

chaoseverlasting

tanA=sinA/cosA.

Alternatively, you could use $$tanA=\frac{u^2sin^2A}{gD}$$

5. Feb 15, 2007

chaoseverlasting

sin2A=2sinAcosA, u=v_0

6. Feb 15, 2007

Aggie

D is the range

7. Feb 16, 2007

Aggie

i found the answer. I used Sin A = gD/Vo^2
Had a little help

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