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 = 12.0 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 y2 = 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?
To find the x and y coordinates of the melon when it splatters on the bank, we can use the equations of motion in two dimensions.
Let's consider the motion of the melon in the x-direction first. Since the initial speed is given as 12.0 m/s, the initial velocity in the x-direction is also 12.0 m/s. The acceleration in the x-direction is 0 because there are no external forces acting in that direction once the melon leaves the truck.
Using the equation of motion in the x-direction:
x = x₀ + v₀_xt + 0.5at²
Since there is no acceleration in the x-direction, the equation simplifies to:
x = x₀ + v₀_xt
Here, x₀ is the initial position in the x-direction, and v₀_x is the initial velocity in the x-direction.
Now let's consider the motion of the melon in the y-direction. The equation of the bank, y² = 16x, can be rearranged to y = √(16x).
Using the equation of motion in the y-direction:
y = y₀ + v₀_yt + 0.5gt²
Here, y₀ is the initial position in the y-direction, v₀_y is the initial velocity in the y-direction, and g is the acceleration due to gravity (-9.8 m/s²).
The initial position in the y-direction is at the edge of the road (y₀ = 0). The initial velocity in the y-direction can be found using the fact that the melon rolls off with an initial horizontal speed of 12.0 m/s. Since there is no initial vertical speed, v₀_y = 0.
Plugging these values into the equation of motion in the y-direction, we get:
y = 0 + 0 + 0.5(-9.8)t²
y = -4.9t²
Now we can solve the two equations simultaneously to find the time (t) at which the melon splatters on the bank.
Substituting the expression for y from the second equation into the first equation, we have:
√(16x) = x₀ + v₀_xt
Substituting the known values into this equation, we get:
√(16x) = 0 + 12.0t
Squaring both sides of the equation, we have:
16x = 144t²
Dividing both sides of the equation by 16, we get:
x = 9t²
Now we have two equations:
y = -4.9t²
x = 9t²
Since both x and y are expressed in terms of t², we can equate them:
-4.9t² = 9t²
Simplifying the equation, we have:
14.9t² = 0
This equation implies that t = 0.
Substituting t = 0 into the equations for x and y, we find that the x-coordinate and y-coordinate of the melon when it splatters on the bank are both 0.
Therefore, the melon splatters on the bank at the origin (0, 0).