A ball is thrown from an initial height of 4 feet with an initial upward velocity of 30 feet per second.

I guess that we want to find the position equation for the ball.

First, we know that any object that is in the air has acceleration equal to the gravitational acceleration (where we ignore forces like air resistance and others) so we can write the acceleration equation as:

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

Where 32 ft/s^2 is the gravitational acceleration, and the negative sign is because this acceleration is downwards.

To get the velocity equation, we can integrate over time to get:

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

where v0 is the initial velocity.

We know that the ball is thrown upwards with a velocity of 30 ft/s, then:

v0 = 30ft/s

And the velocity equation becomes:

v(t) = (-32 ft/s^2)*t + 30 ft/s

Finally, for the position equation we need to integrate again, to get:

p(t) = (1/2)*(-32 ft/s^2)*t^2 + (30 ft/s)*t + p0

where p0 is the initial position.

We know that the ball is thrown from an initial height of 4 ft, then:

p0 = 4ft

Then the position equation (or height equation) of the ball is:

p(t) =  (1/2)*(-32 ft/s^2)*t^2 + (30 ft/s)*t + 4 ft

answer is the third choice:

3x^5+4x^4+x^3+2x^2-10x-7=0

step-by-step explanation: