Question

If the archerfish spits its water 60 degrees from the horizontal aiming at an insect 1.4 m above the surface of the water, how fast must the fish spit the water to hit its target? The insect is at the highest point of the trajectory of the spit water. Use g = 10 m/s2.

Answer 1

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

Gravitational acceleration = g = -10 m/s²

Initial velocity of the water = V₀

Angle at which the water is spit = = 60^o

Initial horizontal velocity of the water = V_x0 = V₀Cos

Initial vertical velocity of the water = V_y0 = V₀Sin

Height at which the insect is = H = 1.4 m

The insect is at the highest point of the trajectory of the spit water.

Vertical velocity of the water when it hits the insect = V_y1 = 0 m/s

V_y1² = V_y0² + 2gH

(0)² = V_y0²+ 2(-10)(1.4)

V_y0 = 5.29 m/s

V_y0 = V₀Sin

5.29 = V₀Sin(60)

V₀ = 6.11 m/s

Speed at which the fish must spit the water to hit its target = 6.11 m/s