# I'm confused on this kinematics question. Can someone help me?

An archerfish hunts by dislodging an unsuspecting insect from its resting place with a stream of water expelled from the fish's mouth. Suppose the archerfish squirts water with a speed of 2.00 m/s at an angle of 50 degrees above the horizontal, and aims for a beetle on a leaf 3.00 cm above the water's surface.

a. Find the time it takes for the water to hit the beetle; why are there 2 answers ?

b. At what horizontal distance from the beetle should the archerfish fire if it is to hit its target in the least time ?

c. Find the angle (above the horizontal) of the archerfish's line of sight; explain why the line of sight angle is not 50 degrees !

### 1 Answer

- RealProLv 71 month ago
What are you confused about?

y(t) = 0 + (2sin50°)t - (9.81/2)t^2

(2sin50°)t - (9.81/2)t^2 = 0.03

Solve for t.

Multiply the smaller solution by the horizontal velocity 2cos(50°) m/s to find required horizontal distance.

Straight line angle = arctan(y / x), obviously lower than 50° but not by much because 2 m/s is fast for such a small distance.

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