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Bullet Motion - help!!!

 
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FloraL



Joined: 06 Jul 2015
Posts: 1
Location: Australia

PostPosted: Tue Jul 07, 2015 8:53 am    Post subject: Bullet Motion - help!!! Reply with quote

[b]A man is standing with a gun and is shooting a target which is 300 m away
The target is 1.65m above the ground
The man is 1.55m from the ground (I think that's his height?)
The bullet-barrel velocity is 550 m/s
How do I calculate the angle above the horizontal of the gun?[/b]

I got an answer for this, but I'm not sure if it is correct, or if the method I used to find the angle is correct, so here's what I did so far:

Initial horizontal velocity: 550cos@
Initial vertical velocity: 550sin@

Formula: s=ut + 1/2 t^2
to get equations:
300=550cos@t (for horizontal) which can also be arranged to t= 300/550cos@

0.1=(550sin@t) - 4.9t^2 (for vertical)

I then substituted the t-value from the horizontal's equation into the vertical velocity's equation, and after rearranging the substituted equation using trig identities, got a final quadratic equation of
1.5tan^2@ - 300tan@ + 1.6 = 0.

I then substituted this equation into the quadratic formula, which gave me two answers: 200, and (5.3 X 10^-3).

I took the tan inverse of each of these equations and got the angles each gave respectively: 89.71, and 0.3.

As the gun can't be fired at an almost 90 degree angle, I eliminated the first answer and used the second angle for my answer, which seemed more practical, as the bullet would only have to climb for a height of 0.1m to reach the target, therefore the angle would have to be rather small.
Is this all correct, or is there another method of working out the angle,please??

Thanks. Very Happy
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