Question

If the archerfish spits its water 60. degrees from the horizontal aiming at an insect 2.0 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

Answer: To hit the target fish must spit with the velocity of 7.30 m/s.

Step-by-step explanation:

Highest point of the trajectory = H = 2 m

Angle at which archer fish spit = θ = 60°

Acceleration due to gravity = g = 10
`m/s^2`

The maximum height of the projectile is given as:

`H=(u^2\sin^2\theta )/(2g)`

`2 m=(u^2\sin^2(60^o))/(2* 10 m/s^2)`

`\sin 60^o=(√(3))/(2)`

`u^2=(2 m* 2* 10 m/s^2* 4)/(3)=53 m^2/s^2`

`u=7.30 m/s`

To hit the target fish must spit with the velocity of 7.30 m/s.