I\'m having difficulty setting this one up. I\'m not sure wich kinematic equatio

Question

I'm having difficulty setting this one up. I'm not sure wich kinematic equation to use and I would apppreciate if anyone could explain how to implement the right kinematic equation or why that equation was used. Answers would also be very helpful since Webassing is particular about rounding errors.

The small archerfish (length 20 to 25 cm) lives in brackish waters of Southeast Asia from India to the Philippines. This aptly named creature captures its prey by shooting a stream of water drops at an insect, either flying or at rest. The bug falls into the water and the fish gobbles it up. The archerfish has high accuracy at distances of 1.2 m to 1.5 m, and it sometimes makes hits at distances up to 3.5 m. A groove in the roof of its mouth, along with a curled tongue, forms a tube that enables the fish to impart high velocity to the water in its mouth when it suddenly closes its gill flaps. Suppose the archerfish shoots at a target that is 2.25 m away, measured along a line at an angle of 30.0° above the horizontal. With what velocity must the water stream be launched if it is not to drop more than 4.70 cm vertically on its path to the target?

Explanation / Answer

Okay, First off, Lets start off with the X direction: Essentially, if we make a right triangle, then the hypotenuse is going to be 2.25m long. that means, that if we apply X=Vcos(30)t, and X=2.25cos(30) (from the right trangle), then t=2.25/V seconds to reach it. That means that in the projectile motion equation, Y=Vsin(30)t-(1/2)gt^2, we can plug that in for t, since it takes that long to reach it. The problem for you, I think, arises when the "not dropping more than 4.7cm" comes up. Since we know from our right triangle that Y=2.25sin(30), we know that the projectile must reach a height of 2.25sin(30). That is what we want Y final to be. Since it can't be any less than 4.7cm of this, we simply take 2.25sin(30)-.047, which gives us 1.078. Therefore, plugging it into the equation we have: 1.078=Vsin(30)t-(1/2)gt^2. Plugging 2.25/v in for t we get: 1.078=Vsin(30)(2.25/V)-(1/2)g(2.25/V)^2 which follows with: 1.078=2.25sin(30)-(1/2)g(2.25)^/V^2. Now, algebraically Isolate for V^2 and solve. 1.078-2.25sin(30)=-(1/2)g(2.25)^2/V^2 -----> 1/V^2=- (2)(1.078-2.25sin(30))/(g*2.25^2) So, V=22.985 m/s

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