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

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:

`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+(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+(-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)