Question

A rock is launched from a cannon. Its height, h(x), can be represented by a quadratic function in terms of time, x, in seconds. After 1 second, the rock is ______ feet in the air. After 2 seconds, it is ______ feet in the air.

Answer 1

Without the specific quadratic function or initial conditions, we cannot calculate the rock's height after 1 or 2 seconds. Normally, this would involve plugging the time values into a quadratic function representing projectile motion, which typically includes gravity's acceleration, initial velocity, and initial height.

Step-by-step explanation:

To answer how high the rock is after 1 and 2 seconds when launched from a cannon, we would need the explicit quadratic function that describes the rock's height over time. However, since we do not have a specific equation, we can discuss the general process instead. The height h(x) at a given time x can typically be found using the standard form of a quadratic equation for projectile motion:
`h(x) = -4.9x^2 + v_0x + h_0,` where -4.9 is half the acceleration due to gravity (in meters per second squared) when working with the metric system, v_0 is the initial velocity in meters per second, and h_0 is the initial height from which the rock is launched.

Without the specific values for initial velocity and starting height in feet, or a conversion factor to translate meters to feet, we cannot calculate the exact heights after 1 and 2 seconds. Typically in the U.S., the acceleration due to gravity would be represented as
`-16 (ft/s^2)` , since gravity accelerates objects at approximately 32 feet per second squared. The concept remains the same, but the student would need to provide the initial velocity and starting height in the appropriate units to obtain an answer.

If we were given these initial conditions and asked to calculate the position and velocity, we would substitute the time values into the quadratic function and solve for h(x) to find the height at each second.