Physics
posted by Leah on .
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 2.45 m/s at an angle of 53.0 degrees above the horizontal, and aims for a beetle on a leaf 4.00 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?
How much time will the beetle have to react?
I tried t=2.45sin(53)/9.81 and x=2.45cosine(53)*t but they both gave me wrong answers? What exactly do I need to do?

To hit in the least time get the target while the water is still rising, not coming back down,
h = .04 meters high
a = g = 9.8 m/s^2
vertical initial speed Vo = 2.45 sin 53 = 1.96 m/s
z = Zo + Vo t +.5 a t^2
.04 = 1.96 t  4.9 t^2
4.9 t^2  1.96 t + .04 = 0
solve quadratic for t, use shortest of the two answers
then horizontal distance = t *2.45 cos 53 
Thx I got it!!!

Wtf. Damon you are horrible at this. this is wrong again

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 2.25m/s at an angle of 48.0 ∘ above the horizontal, and aims for a beetle on a leaf 3.50cm above the water's surface.