How High to Aim Above a Target to Correct for Gravity?

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SUMMARY

The discussion focuses on calculating the angle above a target to account for gravitational drop when shooting a bullet at a velocity of 409 m/s towards a target 1,290 meters away. The equation used is D * Tan A = (g/2) * (D/Vo)^2 * (Cos A)^2, where D is the distance to the target and Vo is the initial velocity. Participants suggest using trigonometric identities to isolate Tan A and derive the necessary angle for accurate targeting. The final solution involves using Sin A = gD/Vo^2 to find the correct angle.

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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
 
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How is your Trig?

Do you know the definition of TanA in terms of SinA and CosA? Use that to isolate A.
 
That's as far as I can go
 
tanA=sinA/cosA.


Alternatively, you could use tanA=\frac{u^2sin^2A}{gD}
 
sin2A=2sinAcosA, u=v_0
 
D is the range
 
i found the answer. I used Sin A = gD/Vo^2
Had a little help
 

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