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Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a spitting cobra rears up to a height of 0.400 m above the ground and launches venom at 2.80 m/s, directed 39.0° above the horizon. Neglecting air resistance, find the horizontal distance (in m) traveled by the venom before it hits the ground.

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Answer: 1.124 m

Explanation:

This situation is a good example of the projectile motion or parabolic motion, and the main equations that will be helpful in this situations are:  

x-component:  

[tex]x=V_{o}cos\theta t[/tex]   (1)  

Where:  

[tex]V_{o}=2.80 m/s[/tex] is the initial speed  

[tex]\theta=39\°[/tex] is the angle at which the venom was shot

[tex]t[/tex] is the time since the venom is shot until it hits the ground  

y-component:  

[tex]y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}[/tex]   (2)  

Where:  

[tex]y_{o}=0.4 m[/tex]  is the initial height of the venom

[tex]y=0[/tex]  is the final height of the venom (when it finally hits the ground)  

[tex]g=-9.8m/s^{2}[/tex]  is the acceleration due gravity

Knowing this, let's begin:

First we have to find [tex]t[/tex] from (2):

[tex]0=0.4 m+2.8 m/s sin(39\°) t+\frac{-9.8m/s^{2}t^{2}}{2}[/tex]   (3)

Rearranging (3):  

[tex]-4.9 m/s^{2} t^{2} + 1.762 m/s t + 0.4 m=0[/tex]   (4)  

This is a quadratic equation (also called equation of the second degree) of the form [tex]at^{2}+bt+c=0[/tex], which can be solved with the following formula:  

[tex]t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}[/tex] (5)  

Where:  

[tex]a=-4.9[/tex]  

[tex]b=1.762[/tex]  

[tex]c=0.4[/tex]  

Substituting the known values:  

[tex]t=\frac{-1.762 \pm \sqrt{(1.762)^{2} - 4(-4.9)(0.4)}}{2(-4.9)}[/tex] (6)  

Solving (6) we find the positive result is:  

[tex]t=0.517 s[/tex] (7)

Substituting (7) in (1):

[tex]x=2.8 m/s cos(39\°) (0.517 s)[/tex]   (8)

Finally:

[tex]x=1.124 m[/tex]   (9)

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