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 0.400 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.

Answer 1

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:  

$x=V_{o}cos\theta t$   (1)  

Where:  

$V_{o}=2.80 m/s$ is the initial speed  

$\theta=39\°$ is the angle at which the venom was shot

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

y-component:  

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

Where:  

$y_{o}=0.4 m$  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

Knowing this, let's begin:

First we have to find $t$ from (2):

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

Rearranging (3):  

$-4.9 m/s^{2} t^{2} + 1.762 m/s t + 0.4 m=0$   (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:  

$a=-4.9$  

$b=1.762$  

$c=0.4$  

Substituting the known values:  

$t=\frac{-1.762 \pm \sqrt{(1.762)^{2} - 4(-4.9)(0.4)}}{2(-4.9)}$ (6)  

Solving (6) we find the positive result is:  

$t=0.517 s$ (7)

Substituting (7) in (1):

$x=2.8 m/s cos(39\°) (0.517 s)$   (8)

Finally:

$x=1.124 m$   (9)