A ball is thrown straight down from the top of a 435-foot building with an initial velocity of -27 feet per second. Use the position function below for free-falling objects. s(t) = -16t^2 + v_0t + s_0 What is its velocity after 2 seconds? v(2) = -91 ft/s What is its velocity after falling 364 feet? v = 1.61 ft/s Find an equation of the parabola y = ax^2 + bx + c that passes through (0, 1) and is tangent to the line y = 5x - 5 at (1, 0). Y = 5x + 10

a) The velocity of the ball after 2 seconds is -91 feet per second, b) The velocity of the ball after falling 364 feet is 155 feet per second, c) The equation of the parabola that passes through (0,1) and is tangent to the line y = 5x - 5 is .

Step-by-step explanation:

a) The velocity function is obtained after deriving the position function in time:

The velocity of the ball after 2 seconds is:

The velocity of the ball after 2 seconds is -91 feet per second.

b) The time of the ball after falling 364 feet is found after solving the position function as follows:

The solution of this second-grade polynomial is represented by two roots:

and .

Only the first root is physically reasonable since time is a positive variable. Now, the velocity of the ball after falling 364 feet is:

The velocity of the ball after falling 364 feet is 155 feet per second.

c) Let consider the equation for a second order polynomial that passes through (0, 1) and its first derivative that passes through (1, 0) and represents the give equation of the tangent line. That is to say:

Second-order polynomial evaluated at (0, 1)

Slope of the tangent line evaluated at (1, 0)

Now, let evaluate the second order polynomial at (1, 0):

If and , then:

And the value of b is: ()

The equation of the parabola that passes through (0,1) and is tangent to the line y = 5x - 5 is .

the function makes a strait line horizontally with no x intercept but a y intercept at 9.

actually your concept is wrong

step-by-step explanation:

150% of x =300

150x=300×100

x=(300×100)÷150

x=200