The archerfish hunts by dislodging an unsuspecting insect from its resting place

Question

The 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 2.05 m/s at an angle of 55.0 above the horizontal, and aims for a beetle on a leaf 4.00 cm above the water's surface.

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

Part B) How much time will the beetle have to react?

Thank you for helping!

Explanation / Answer

(a)
Horizontal speed, vx = 2.05 * cos(55.0) = 1.175 m/s
Vertical speed, vy = 2.05 * sin(55.0) = 1.679 m/s

Let Time taken to reach beetle = t
Let the Horizontal distance be x.

t = x/1.175 -------1

0.04 = 1.679*t - 1/2*9.8*t^2
0.04 = 1.679 * (x/1.175) - 4.9 * (x/1.175)^2
x = 0.0302 m

Horizontal distance, x = 3.02 cm

(b)
Time beetle will have to react,
t = 0.0302/1.175
t = 0.0257 s

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