Suppose an object is launched from ground level directly upward at 57.4 f/s Write a function to represent the object’s height over time.

p(t) = (-16 ft/s^2)*t^2 + (57.4 ft/s)*t

Step-by-step explanation:

We can suppose that the only force acting on the object is the gravitational force, then the acceleration of the object will be equal to the gravitational acceleration.

Then we can write:

a(t) = -32 ft/s^2

Where the negative sign is because this acceleration is downwards.

Now, to get the vertical velocity of the object, we need to integrate over time to get:

v(t) = (-32 ft/s^2)*t + v0

where t represents time in seconds and v0 is the constant of integration, and in this case, is the initial vertical velocity.

In this case, the initial velocity is 57.4 ft/s upwards, then the velocity equation is:

v(t) = (-32 ft/s^2)*t + 57.4 ft/s

To get the position equation we need to integrate over time again, to get:

p(t) = (1/2)*(-32 ft/s^2)*t^2 + (57.4 ft/s)*t + p0

Where p0 is the initial height of the object, as it was launched from the ground, then the initial position is p0 = 0ft.

then the position equation (that is the function that represents the height of the object as a function over time) is:

p(t) = (1/2)*(-32 ft/s^2)*t^2 + (57.4 ft/s)*t

p(t) = (-16 ft/s^2)*t^2 + (57.4 ft/s)*t

the answer is (4,3)

since the rule for r0, 90º is (-y,x) and it was already applied to form the image, all you have to do is reverse the answer using this formula.

step-by-step explanation: