Solving a Trajectory Problem Involving a Watermelon

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To solve the trajectory problem involving the watermelon, start by visualizing the scenario with a diagram to understand the melon’s path. The initial horizontal velocity is 10.0 m/s, and the equation of the bank is y^2 = 16x, which describes a parabolic shape. Analyzing the motion, the horizontal acceleration is zero while the vertical acceleration is -g (gravity). The key is to determine the intersection of the watermelon’s trajectory with the parabolic bank to find the coordinates where it lands. Understanding the ground's shape is crucial for calculating the exact landing point of the watermelon.
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Homework Statement


A truck loaded with cannonball watermelons stops suddenly to avoid running over the edge of a washed-out bridge (see figure). The quick stop causes a number of melons to fly off the truck. One melon rolls over the edge with an initial speed vi = 10.0 m/s in the horizontal direction. A cross-section of the bank has the shape of the bottom half of a parabola with its vertex at the edge of the road, and with the equation y^2= 16x, where x and y are measured in meters. What are the x and y coordinates of the melon when it splatters on the bank?

How would u start to solve this?


Homework Equations


vi=10.0m/s y^2=16x


The Attempt at a Solution


umm...ax=0,ay=-g
 
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If I were you, I’d start off by drawing a picture.

<br /> y^2 = 16x <br />

isn’t very helpful in this instance, how could you get a better picture of the ground. Once you know what the ground looks like it becomes easier to see where the path of the watermelon intersects the cliff.

How would you find out then where it lands?
 
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