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/s.

6.1 m/s

Explanation:

Take up to be positive.  Given (in the y direction):

Δy = 1.4 m

vᵧ = 0 m/s

g = -10 m/s²

Find: v₀ᵧ

vᵧ² = v₀ᵧ² + 2aΔy

(0 m/s)² = v₀ᵧ² + 2 (-10 m/s²) (1.4 m)

v₀ᵧ = 5.29 m/s

The vertical component is 5.29 m/s.  So the total magnitude of the initial velocity is:

v₀ᵧ = v₀ sin θ

5.29 m/s = v₀ sin 60°

v₀ = 6.11 m/s

Rounding to two significant figures, the fish must spit the water at 6.1 m/s.

float.

archimedes' principle

denser than water sinks in water.

denser than mercury would sink in it

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//