Question

There is another equation that describes the path of your fruit or vegetable. This equation is: y=-16t^2+v0t+6 .

• y is the height of the object.
• t is the time, in seconds, since it was launched,
• v0 is a constant. It's the initial upward velocity (speed) of the object, in feet per second
• -16 accounts for the pull of gravity.
• 6 means the object was 6 feet off the ground when launched.

11. How is this equation different from the one you wrote in question 9? How are the equations similar?

The equation in question 9 is y=-0.0992x^2+4.45x

Answer 1

The equations have the same mathematical form:

S = A + V0 t - 1/2 g t^2

They are both quadratic equations in one variable (t here)

This equation expresses the distance traveled by an object in time t in the vertical direction - it does not necessarily refer to a graph

In the other equation, y can be considered a vertical coordinate equation on a graph and x would be the horizontal coordinate

Note that in one equation 3 terms are necessary to describe the displacement of an object in one direction

In the other 2 terms in the x-direction describe the displacement in the other or y-direction

Answer 2

The equation provided describes projectile motion, where an object is launched and impacted on a horizontal surface. Physics principles are applied to calculate various aspects of motion, like the apex of the trajectory and time of flight, which depend on the initial velocity and gravity.

The equation y = -16 t^2 + v_0 t + 6 describes the vertical position of a projectile over time under the influence of gravity. This type of motion is a common topic in physics, specifically when examining the motion of an object thrown into the air and returning back to the same height from which it was launched.

The term -16 represents the acceleration due to gravity in feet per second squared (assuming the value of g is -32 ft/s², and the equation is adjusted to feet rather than meters).

At the highest point, the vertical velocity (vy) becomes 0, which is the apex of the trajectory. To find this point, we can use the equation v² = v₀² - 2g(y - y₀), where v is the final velocity, v₀ is the initial velocity, g is the acceleration due to gravity, and y and y₀ are the final and initial heights respectively.

For a projectile launched and landing at the same elevation, the time of flight is proportional to the initial upward velocity and inversely proportional to gravity.

Considering the equation y = y₀ + (v₀y + vy)t, where t is the time, v₀y is the initial vertical velocity, and g is again the acceleration due to gravity, this reflects the vertical component of the projectile's motion.

To determine the time at the highest point (when vy = 0), one would solve for t using the simplified equation: y = (v₀y + vy)t, where y₀ is zero.