Answer: 1.289 m
Explanation:
The path the cobra's venom follows since it is spitted until it hits the ground, is described by a parabola. Hence, the equations for parabolic motion (which has two components) can be applied to solve this problem:
x-component:
[tex]x=V_{o}cos\theta t[/tex] (1)
Where:
[tex]x[/tex] is the horizontal distance traveled by the venom
[tex]V_{o}=3.10 m/s[/tex] is the venom's initial speed
[tex]\theta=47\°[/tex] is the angle
[tex]t[/tex] is the time since the venom is spitted 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}=0.44 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
Let's begin with (2) to find the time it takes the complete path:
[tex]0=0.44 m+3.10 m/s sin\theta(47\°)+\frac{-9.8m/s^{2} t^{2}}{2}[/tex] (3)
Rewritting (3):
[tex]-4.9 m/s^{2} t^{2} + 2.267 m/s t + 0.44 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 m/s^{2[/tex]
[tex]b=2.267 m/s[/tex]
[tex]c=0.44 m[/tex]
Substituting the known values:
[tex]t=\frac{-2.267 \pm \sqrt{2.267^{2}-4(-4.9)(0.44)}}{2(-4.9)}[/tex] (6)
Solving (6) we find the positive result is:
[tex]t=0.609 s[/tex] (7)
Substituting (7) in (1):
[tex]x=(3.10 m/s)cos(47\°)(0.609 s)[/tex] (8)
We finally find the horizontal distance traveled by the venom:
[tex]x=1.289 m[/tex]
