Part a at what speed v should an archerfish spit the water to shoot down an insect floating on the water surface located at a distance 0.800 m from the fish? assume that the fish is located very close to the surface of the pond and spits the water at an angle 60∘ above the water surface.

3.01 m/s

This is a simple projectile calculation. What we want is a vertical velocity such that the time the droplet spends going up and going back down to the surface exactly matches the time the droplet takes to travel horizontally 0.800 meters. The time the droplet spends in the air will is:

V*sqrt (3) / 2; Vertical velocity.

(V*sqrt (3) / 2) / 9.8; Time until droplet reaches maximum height

(V*sqrt (3)) / 9.8; Double that time for droplet to fall back to the surface.

The droplet's horizontal velocity will be:

V/2.

So the total distance the droplet travels will be:

d = (V*sqrt (3)) / 9.8 * V/2

d = V^2*sqrt (3) / 19.6

Let's substitute the desired distance and solve for V

d = V^2*sqrt (3) / 19.6

0.8 = V^2*sqrt (3) / 19.6

15.68 = V^2*sqrt (3)

15.68/sqrt (3) = V^2

15.68/1.732050808 = V^2

3.008795809 = V

So after rounding to 3 significant figures, the archerfish needs to spit the water at a velocity of 3.01 m/s

Let's verify that answer.

Vertical velocity: 3.01 * sin (60) = 3.01 * 0.866025404 = 2.606736465

Time of flight = 2.606736465 * 2 / 9.8 = 0.531987034 seconds.

Horizontal velocity: 3.01 * cos (60) = 3.01 * 0.5 =