   Chapter 3, Problem 47AP

Chapter
Section
Textbook Problem

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 0.500 m above the ground and launches venom at 3.50 m/s, directed 50.0° above the horizon. Neglecting air resistance, find the horizontal distance traveled by the venom before it hits the ground.

To determine
The horizontal distance travelled by the venom before it hits the ground.

Explanation

The components of the initial velocity of the venom when v0=3.50m/s and θ=50.0° are,

v0x=v0cosθ

Here,

v0 is the initial velocity

θ is the angle of projection

v0x=(3.50m/s)cos50.0°=2.25m/s

v0y=v0sinθ=(3.50m/s)sin50.0°=2.68m/s

The final velocity of the venom before hitting the ground is negative and is given by

vy=voy22gΔy

Here,

vy is the vertical velocity

v0y is the vertical component of the initial velocity

g is the acceleration due to gravity

Δy is the vertical displacement

Substitute 2.68m/s for v0y , 0.500m for Δy and 9.80m/s2 for g .

vy=(2

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