Gunshot Projectile Motion: Two possible angles*

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SUMMARY

The discussion focuses on calculating the two possible angles (θ1 and θ2) for a bullet fired from a rifle with a muzzle speed of 424 m/s to hit a target 91.4 meters away. The relevant equations of motion include the horizontal distance formula and the trigonometric identity 2 sin(θ) cos(θ) = sin(2θ). The calculated time for the bullet to reach the target is 0.2219 seconds, which is essential for determining the angles using projectile motion principles.

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Homework Statement


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 = 424 m/s, there 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. Give your answers as (a) the smaller angle and (b) the larger angle. (Hint: The following trigonometric identity may be useful: 2 sinθ cosθ = sin 2θ.)


Homework Equations


2sin(theta)cos(theta) = sin2(theta)


The Attempt at a Solution



acceleration in the y direction = -9.80 m/s^2
initial velocity = 424 m/s
distance in the x direction = 94.1 m
calculated time.
94.1 m/424 m t = 0.2219 seconds

I have two triangles drawn (representing the extreme and normal theta). I am unsure of how to calculate the velocity in the x direction so I can use trig to calculate my angle and y component of the triangle. Solving one triangle will help me solve the second one.

Thanks.
 
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You do not know that the time taken is 0.2219 seconds.

What are the horizontal and vertical equations of motion for the bullet? Those are the relevant equations here.
 

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