Question

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 0.440 m above the ground and launches venom at 3.10 m/s, directed 47.0° above the horizon. Neglecting air resistance, find the horizontal distance (in m) traveled by the venom before it hits the ground.

Answer 1

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:

$x=V_{o}cos\theta t$  (1)

Where:

$x$ is the horizontal distance traveled by the venom

$V_{o}=3.10 m/s$ is the venom's initial speed

$\theta=47\°$ is the angle

$t$ is the time since the venom is spitted until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$   (2)

Where:

$y_{o}=0.44 m$  is the initial height of the venom

$y=0$  is the final height of the venom (when it finally hits the ground)

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

Let's begin with (2) to find the time it takes the complete path:

$0=0.44 m+3.10 m/s sin\theta(47\°)+\frac{-9.8m/s^{2} t^{2}}{2}$   (3)

Rewritting (3):

$-4.9 m/s^{2} t^{2} + 2.267 m/s t + 0.44 m=0$   (4)

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

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (5)

Where:

$a=-4.9 m/s^{2$

$b=2.267 m/s$

$c=0.44 m$

Substituting the known values:

$t=\frac{-2.267 \pm \sqrt{2.267^{2}-4(-4.9)(0.44)}}{2(-4.9)}$ (6)

Solving (6) we find the positive result is:

$t=0.609 s$ (7)

Substituting (7) in (1):

$x=(3.10 m/s)cos(47\°)(0.609 s)$  (8)

We finally find the horizontal distance traveled by the venom:

$x=1.289 m$