**47. ssm The drawing shows an exaggerated view of a rifle that has been “sighted in” for a 91.4-meter target. If the muzzle speed of the bullet is v0 = 427 m/s, what are the two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target? One of these angles is so large that it is never used in target shooting. (Hint: The following trigonometric identity may be useful: 2sin(θ)cos(θ) = sin(2θ).)

 

 

Horizontal motion gives us

 

 

We get time from the y motion which has a net zero displacement as we have the bullet back at the same height it was launched at.

 

 

 

Again a product results in zero means either t = 0 or

 

Solving for t

Plug that into our distance equation for x

 

 

Using the trig identity provided we get

 

Solve for 2q

 

 

 

Now complimentary angles add to make 90°, so this actually gives two possible angles

 

 

 is not reasonable as you would be firing straight over head and that would be dangerous, so the answer must be

 

 

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This page last updated on January 11, 2020