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 0.420 m above the ground and launches venom at 2.50 m/s, directed 35.0° above the horizon. Neglecting air resistance, find the horizontal distance (in m) traveled by the venom before it hits the ground.

Answer 1

The horizontal distance traveled by the venom before it hits the ground is 0.6 m

What is Projectile ?

A stone or ball or anything projected is known as a projectile.

Given that  a spitting cobra rears up to a height of 0.420 m above the ground and launches venom at 2.50 m/s, directed 35.0° above the horizon.

Assuming air resistance is neglected, the parameters to be considered are;

• Height h = 0.42 m

• Velocity v = 2.5 m/s

• Angle Ф = 35°

The ball we reach the ground at the same time when it is dropped vertically or horizontally.

In a vertical direction,

h = 1/2g$t^{2}$

Substitute all the parameters into the formula

0.42 = 1/2 x 9.8 x $t^{2}$

0.42 = 4.9$t^{2}$

$t^{2}$ = 0.42/4.9

$t^{2}$ = 0.0857

t = $\sqrt{0.0857}$

t = 0.29 s

The horizontal distance traveled by the venom before it hits the ground will be

Distance = ucosФ x t

Distance = 2.5cos35 x 0.29

Distance = 0.599 m

Distance = 0.6 m

Therefore,  the horizontal distance (in m) traveled by the venom before it hits the ground is 0.6 m

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