Question

Spitting cobras can defend themselves by squeezing muscles around the venom glands to squirt venom at an attacker. Supposed spitting cobra rears up to a height of 0.500 m above the ground and launches venom at 3.50 m/s, directed at 50.0° above the horizon. neglecting air is a stance, find the horizontal distance traveled by the venom before it hits the ground.

Answer 1

Answer: 1.56 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:

$x$ is the horizontal distance

$V_{o}=3.5 m/s$ is the initial speed

$\theta=50\°$ 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:

$y_{o}=0.5m$  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):

$0=0.5 m+3.5m/s sin(50\°) t+\frac{-9.8m/s^{2}t^{2}}{2}$   (3)

Rearranging (3):

$-4.9 m/s^{2} t^{2} + 2.681 m/s t + 0.5 m=0$   (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:

$a=-4.9$

$b=2.681$

$c=0.5$

Substituting the known values:

$t=\frac{-2.681 \pm \sqrt{(2.681)^{2}-4(-4.9)(0.5)}}{2(-4.9)}$ (6)

Solving (6) we find the positive result is:

$t=0.694 s$ (7)

Substituting (7) in (1):

$x=3.5 m/s cos(50\°) (0.694 s)$   (8)

Finally:

$x=1.56 m$   (9)