The archerfish hunts by dislodging an unsuspecting insect from its resting place with a stream of water expelled from the fish's mouth.

Question

Suppose the archerfish squirts water with a speed of 2.50 m/s at an angle of 56.0 ∘ above the horizontal, and aims for a beetle on a leaf 3.50 cm above the water's surface.

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

2) How much time will the beetle have to react?

Answer 1

Solution:

The formula of the distance required for a projectile to reach its target is X = V₀ cos α t

The formula for the maximum height is y_max = -1/2gt²+V_osinα t + y₀

The initial height is the surface of the water y₀ = 0 , the Maximum height is y_max = 3.5cm=0.035m or the height of the beetle. Substituting 0.035m =-1/2(9.81m/s²)(t²)+2.5m/s sin 56° t + 0

0.035m = -4.91 t² + 2.073t . Rearranging you get 4.91t² -2.073 t + 0.035 = 0

solving the quadratic equation divide all terms by 4.91, t² - 0.422t + 0.00713 =0

t = 0.422 ± √[(-0.422)² -4(1)(0.0173)] / 2 = 0.422 ± √0.15 / 2 = 0.422 ± 0.387 / 2

t = 0.404 s = 0.40 s , and t = 0.0175. The latter answer for the time needed is too small so neglect

The required distance X = 2.5m/s cos 56° ( 0.40 s) = 0.559 m = 0.6 m

The beetle has time t to react = 0.4 s