The archerfish hunts by dislodging an unsuspecting insect fr

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 2.30 m/s at an angle of 56.0 above the horizontal, and aims for a beetle on a leaf 3.00 cm above the water\'s surface. At what horizontal distance from the beetle should the archerfish fire if it is to hit its target in the least time?

Solution

vo = 2.30 m/s
theta = 56.0 deg
y = 3.000 cm = 0.030 m
x = ?

y = vo sin(theta) t - 1/2 * gt^2
0.030 = 2.30 sin(56.0 deg) t - 1/2 * 9.8 * t^2
0.030 = 2.30 sin(56.0 deg) t - 1/2 * 9.8 * t^2
0.030 = 1.9068 t - 4.9 t^2
4.9 t^2 - 1.9068 t + 0.030 = 0
t =
t = 0.016426 s, 0.372716 s
Minimum value of t = 0.016426 s
Maximum value of x = vo cos(theta) * t
= 2.30 cos(56.0 deg) * 0.016426
= 0.021 m = 2.1 cm