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

Using the kinematic equations below, we can calculate the initial velocity required. Angle of projectile = 60 degrees Acceleration due to gravity (Ay) = -10 m/s^2 (negative because downward) Height of projectile (Dy) = 2m Vfy^2=Voy^2 +2*Ay*Dy Vfy = 0 m/s because the vertical velocity slows to zero at the height of its trajection. So... 0 = Voy^2 + 2(-10)(2) 0 = Voy^2 - 40 40 = Voy^2 Sqrt40 = Voy 6.32 m/s = Voy THIS IS NOT THE ANSWER. THIS IS JUST THE INITIAL VELOCITY IN THE Y DIRECTION. Using trigonometry, Tan 60 = Voy/Vox. Tan 60 = 6.32/Vox. Vox*Tan 60 = Vox Vox = 10.95 m/s. Now, using Vox = 10.95 and Voy = 6.32, we can use pythagorean theorem to find the total Vo. A^2 +B^2 = C^2 10.95^2 + 6.32^2 = C^2 Solving for C = 12.64 m/s This is the velocity required to hit the surface. You can also calculate a bunch of other stuff now using the other kinematic equations. V = 12.64 m/s

Answer 2

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

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:

$H=\frac{u^2\sin^2\theta }{2g}$

$2 m=\frac{u^2\sin^2(60^o)}{2\times 10 m/s^2}$

$\sin 60^o=\frac{\sqrt{3}}{2}$

$u^2=\frac{2 m\times 2\times 10 m/s^2\times 4}{3}=53 m^2/s^2$

$u=7.30 m/s$

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