Solving a Trajectory Problem Involving a Watermelon

[MEAHH]

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

 

[RyanSchw]

If I were you, I’d start off by drawing a picture.

$$y^2 = 16x$$

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?

 

[Moetasim]

To solve this trajectory problem involving a watermelon, we will first need to understand the initial conditions and the forces acting on the watermelon. From the given information, we know that the watermelon was initially traveling at a horizontal speed of 10.0 m/s and it was subjected to a sudden stop, causing it to fly off the truck. This means that the only force acting on the watermelon after it left the truck is the force of gravity.

To solve this problem, we can use the equations of motion for a projectile in two dimensions, which are:

x = x0 + v0x*t + 0.5*ax*t^2
y = y0 + v0y*t + 0.5*ay*t^2

Where x0 and y0 are the initial positions, v0x and v0y are the initial velocities in the x and y directions respectively, and ax and ay are the acceleration in the x and y directions respectively.

In this case, we can assume that the initial position of the watermelon is at the edge of the road, which corresponds to x0 = 0 and y0 = 0. The initial velocity in the x direction is v0x = 10.0 m/s, and in the y direction, v0y = 0. The acceleration in the x direction, ax, is equal to 0 since there is no force acting in that direction. The acceleration in the y direction, ay, is equal to -g, where g is the acceleration due to gravity (9.8 m/s^2).

We can now substitute these values into the equations of motion and solve for the time it takes for the watermelon to hit the bank. Once we have the time, we can plug it back into the equations to find the x and y coordinates of the watermelon at that time.

It is also important to note that the shape of the bank is given by the equation y^2 = 16x. This means that the y coordinate of the watermelon at any given x coordinate can be found by substituting the value of x into the equation and taking the square root.

In conclusion, to solve this trajectory problem involving a watermelon, we need to use the equations of motion for a projectile and the given information about the initial conditions and the shape of the bank. This will allow us to find the x and y coordinates of the watermelon