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~christina~
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Homework Statement
A Sharp Shooter fish is a fish that hunts insect from the water. It is able to project water droplets toward insects causing them to drop into the water where they can be eaten. In the figure below the sharp shooter is at rest and spots a insect at rest on a leaf hanging over the water. The sharp shooter is 2.2 m away from the bug when it fires a droplet with an initial velocity of 5.98m/s at an agle of 65 degrees relative to the surface of the water. The droplet strikes the bug causing it to drop vertically downward onto the surface of the water.
a) find the maximum height that the water droplet reaches and the time required to reach this height
b) how long will it take the droplet to reach the bug ater leaving the fish's mouth? (ignore air resistance)
c) the fish wants to claim it's prey immediately so that no other fish can steal it's meal. What must the minimum acceleration of the fish be if it is to arrive at the point directly beneath the bug at the same time the bug strikes the surface of the water?
d) sketch graphs of the sharp shooter's displacement, velocity, and acceleration vs time for the time interval required for the fish to claim it's prey.
http://img103.imageshack.us/img103/2894/lastscandn4.th.jpg
Homework Equations
kinematic equations?? I don't know which ones to use...
for the first part...
h= [vi^2 sin^2(theta)]/ 2g
t= (vi sin (theta))/ g
The Attempt at a Solution
I know that :
vi= 5.98m/s
theta= 65 deg
dx= 2.2m
g= 9.8m/s^2
a) max height reached and time it takes
h= [vi^2 sin^2(theta)]/ 2g
h= [(5.98m/s)^2 * sin^2(65)]/ 2*(9.8m/s^2) = 1.498=> 1.50m
t= (vi sin (theta))/ g
t= [(5.98m/s)*sin(65)]/9.8m/s^2 = .553sec
b) I have no idea how to get this
c)I don't know about this either
d) I can't do that unless I find the previous either
I don't know which kinematic equation to use to find the distance..how would I incorperate the angle into the equation if I need to...which I'm not sure about that either.
Thanks
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