A truck loaded with cannonball watermelons stops suddenly to avoid

running over the edge of a washed-out bridge. The quick stop causes a
number of melons to fly off the truck. One melon rolls over the edge with an
initial speed of vi = 12.1m/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 y2 = 14.9x, where x and y are measured
in meters. What is the x coordinate of the melon when it splatters on the bank?
Tries 0/10 What is the y coordinate of the melon when it splatters on the bank?

Well, it seems like this melon really went on an adventure! But don't worry, I'm here to help you solve this fruity mystery.

To find the x-coordinate of the melon when it splatters on the bank, we need to find where the melon intersects with the bank's parabolic shape.

Using the equation y^2 = 14.9x, we can plug in the given values to find the x-coordinate. However, since we're only given the initial horizontal speed of the melon, we don't have enough information to determine how long the melon was in the air before hitting the bank. Therefore, we'll need to make some assumptions.

Let's assume the melon was in the air for a short enough time that we can neglect any horizontal acceleration. In that case, the initial horizontal velocity will remain constant throughout its flight.

Using the equation for horizontal motion, x = vi * t, we can set x equal to the x-coordinate of the bank. Plugging in the initial horizontal speed of the melon, vi = 12.1 m/s, we have:

x = 12.1m/s * t

Now, let's find the y-coordinate of the melon when it splatters on the bank.

Since the melon is in free fall vertically, we can use the equation y = 0.5 * g * t^2, where g is the acceleration due to gravity. Upon impact with the bank, the melon splatters, so its y-coordinate is zero.

0 = 0.5 * 9.8 m/s^2 * t^2

Simplifying this equation, we get:

0 = 4.9m/s^2 * t^2

Now, let's solve for t in this equation. Taking the square root of both sides gives us:

t = 0s or t = sqrt(0)

Since t = sqrt(0) = 0, we can conclude that the melon splatters on the bank instantaneously after falling.

Therefore, the x-coordinate of the melon when it splatters on the bank is x = 12.1m/s * 0s = 0m, and the y-coordinate is y = 0m.

So, the melon splatters right at the edge of the bank, making quite the messy situation for anyone who needs to clean it up!

To find the x-coordinate and y-coordinate of the melon when it splatters on the bank, we need to find the point where the melon's trajectory intersects with the bank.

The equation of the bank is given as y^2 = 14.9x.

We know that the initial horizontal velocity of the melon is vi = 12.1 m/s. Since there is no horizontal acceleration, the horizontal velocity remains constant throughout the motion.

The horizontal distance traveled by the melon can be found using the equation: x = vit, where t is the time.

To find the time it takes for the melon to hit the bank, we need to find the vertical distance traveled by the melon.

At the point where the melon hits the bank, its vertical position is y. The vertical acceleration is due to gravity, and it is -9.8 m/s^2 (negative because it acts downward).

We can use the equation of motion for vertical motion: y = yi + vit + (1/2)at^2, where yi = 0 (melon starts from the top of the bank), vi = 0 (initial vertical velocity), and a = -9.8 m/s^2.

Substituting the values, we get: y = 0 + 0*t - (1/2)*9.8*t^2.

Setting this equation equal to the equation of the bank (y^2 = 14.9x), we have:

(1/2)*9.8*t^2 = 14.9x

Simplifying this equation, we get: 4.9*t^2 = 14.9x.

Now, we can substitute the value of x obtained from the horizontal motion equation into this equation: x = vit.

So, 4.9*t^2 = 14.9(vi*t).

Substituting the values, we have: 4.9*t^2 = 14.9(12.1*t).

Simplifying this equation, we get: 4.9*t^2 = 179.29*t.

Dividing both sides of the equation by t, we get: 4.9*t = 179.29.

Solving for t, we find: t = 179.29 / 4.9 ≈ 36.63 seconds.

Now, we can substitute this value of t back into the equation x = vit to find the horizontal distance traveled by the melon:

x = 12.1 * 36.63 ≈ 442.44 meters.

Therefore, the x-coordinate of the melon when it splatters on the bank is approximately 442.44 meters.

To find the y-coordinate, we can substitute this value of x into the equation of the bank:

y^2 = 14.9 * 442.44.

Simplifying this equation, we get: y^2 ≈ 6585.556.

Taking the square root of both sides, we get: y ≈ √6585.556.

Therefore, the y-coordinate of the melon when it splatters on the bank is approximately √6585.556 meters.

To find the x and y coordinates of the melon when it splatters on the bank, we need to analyze the projectile motion of the melon.

Let's break down the problem into parts:

1. Horizontal motion: Since the initial speed of the melon in the horizontal direction is 12.1 m/s and there is no horizontal acceleration, the horizontal velocity remains constant throughout the motion. Therefore, the horizontal component of the motion does not affect the x coordinate of the melon when it splatters on the bank.

2. Vertical motion: The melon experiences a vertical acceleration due to gravity (g). The equation of motion in the vertical direction is given by y = yo + vot + (1/2)gt², where y is the vertical displacement, yo is the initial vertical position, vo is the initial vertical velocity, t is time, and g is the acceleration due to gravity (approximately 9.8 m/s²).

3. Bank's equation: The equation of the bank's bottom half of the parabola is y² = 14.9x.

Now, let's find the coordinates of the splattered melon:

1. Finding the x coordinate:
Since there is no horizontal acceleration, the x coordinate of the melon remains the same throughout the motion. Therefore, the x coordinate of the melon when it splatters on the bank is the same as its initial x coordinate, which is not specified in the problem statement.

2. Finding the y coordinate:
To find the y coordinate, we need to equate the equation of motion to the equation of the bank's parabola:

(yo + vot + (1/2)gt²)² = 14.9x

Substituting the initial conditions:
(0 + 0 + (1/2)(9.8)t²)² = 14.9x

(4.9t²)² = 14.9x

Since we know the initial vertical speed (vi = 0) of the melon, the time it takes for the melon to hit the bank can be found using the vertical equation:

0 = yo + vot + (1/2)gt²

0 = 0 + 0 + (1/2)(-9.8)t²

t = 0 or t = sqrt(-2(-9.8)/-9.8)

Since time cannot be negative, we'll take t = sqrt(2(-yo)/-g)

Now, substitute this value of t into the equation we found earlier:

(4.9(sqrt(2(-yo)/-g))²)² = 14.9x

98(-yo/g)² = 14.9x

From the given equation of the bank's parabola, we can equate y² = 14.9x:

14.9x = 14.9x

This means the x coordinate is not dependent on the initial vertical position (yo), so we cannot determine the exact value of the x coordinate without knowing the initial position.

To summarize:

- The x coordinate of the melon remains the same throughout the motion.
- The y coordinate of the melon when it splatters on the bank depends on its initial vertical position (yo), which is not given in the problem statement. Therefore, we cannot determine the exact value of the y coordinate without knowing the initial position.

If so then

melon trajectory
u = 12.1 forever
x = 12.1 t so t = x/12.1
v = g t = 9.81 t
y = 4.9 t^2
so
y = 4.9 x^2/146 = .0335 x^2
y^2 = .00112 x^4
when they hit
14.9 x = .00112 x^4
x^3 = 13303
x = 23.7
y = .0335 x^2 = 18.8