- #1

- 21

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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

- Thread starter Aggie
- Start date

- #1

- 21

- 0

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

Integral

Staff Emeritus

Science Advisor

Gold Member

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

- #3

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That's as far as I can go

- #4

- 1,037

- 3

tanA=sinA/cosA.

Alternatively, you could use [tex]tanA=\frac{u^2sin^2A}{gD}[/tex]

Alternatively, you could use [tex]tanA=\frac{u^2sin^2A}{gD}[/tex]

- #5

- 1,037

- 3

sin2A=2sinAcosA, u=v_0

- #6

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D is the range

- #7

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i found the answer. I used Sin A = gD/Vo^2

Had a little help

Had a little help

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