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 0.400 m above the ground and launches venom at 4.00 m/s, directed 59.0° above the horizon. Neglecting air resistance, find the horizontal distance (in m) traveled by the venom before it hits the ground.

Answer 1

Let us consider the upwards direction as positive and the downwards direction as negative.

Gravitational acceleration = g = -9.81 m/s²

Initial velocity of the venom = V = 4 m/s

Angle at which the venom is spit = = 59^o

Initial horizontal velocity of the venom = V_x = VCos = (4)Cos(59) = 2.06 m/s

Initial vertical velocity of the venom = V_y = VSin = (4)Sin(59) = 3.43 m/s

Height from which the cobra spits the venom = H = 0.4 m

Time taken by the venom to reach the ground = T

When the venom lands on the ground the vertical displacement of the venom is in the downwards direction therefore it is negative.

-H = V_yT + gT²/2

-0.4 = (3.43)T + (-9.81)T²/2

4.905T² - 3.43T - 0.4 = 0

T = 0.8 sec or -0.1 sec

Time cannot be negative.

T = 0.8 sec

Horizontal distance traveled by the venom = R

There is no horizontal force acting on the venom therefore the horizontal velocity of the venom remains constant.

R = V_xT

R = (2.06)(0.8)

R = 1.65 m

Horizontal distance traveled by the venom before it hits the ground = 1.65 m