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 abo…

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asked Jan 20, 2020 156k views

1 Answer

Heike · Answer 1 · 2020-01-26T06:04:25+0000

Answer: 1.124 m

Step-by-step 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

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

y-component:

$y=y_(o)+V_(o)sin\theta t+(gt^(2))/(2)$ (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
from (2):

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