What is the biggest insect?

What θ0 is required for the drop to be at the top of the parabolic path when it reaches the insect? ?

• Upon spotting an insect on a twig overhanging water, an archer fish squirts water drops at the insect to knock it into the water. Although the fish sees the insect along a straight-line path at angle φ and distance d, a drop must be launched at a different angle θ0 if its parabolic path is to intersect the insect. If φ = 38.0° and d = 0.900 m what θ0 is required for the drop to be at the top of the parabolic path when it reaches the insect? Figure: * insect on twig / / / d / / / / φ ______________________/_______________… x=o> archer fish

• Answer:

  Since the water droplets are at the top of their trajectory you can write: h = (1/2)gt^2 h = height of insect = 0.9sin(38) = 0.554 m g = acceleration of gravity = 9.8 m/s^2 t = time to fall from the branch to the water t^2 = 2h/g t = SQRT(2h/g) t = SQRT(2*0.554/9.8) = 0.3362 seconds Although this is the time to fall from the branch to the water it is also the time it takes to go from the water to the branch and this we can use below. In order to find the angle phi (P) we can use the fact that tan(P) = v/u where: v = initial vertical velocity of the water u = initial horizontal velocity of the water As far as u it is the horizontal distance divided by the time just calculated so: horizontal distance = 0.9cos(38) = 0.7092 m u = 0.7092/0.3362 = 2.11 m/s We now need v. For this use: v = gt = (9.8)(0.3362) = 3.295 m/s tan(P) = v/u = 3.295/2.11 = 56.37 degrees