A ball is shot from a cannon into the air with an upward velocity of 65 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= -14t2+65t+1.75.
You must show your work.
● Find the maximum height attained by the ball.
● What time was the maximum height achieved?.
● How high was the ball off of the ground when it was shot?
● How far did the ball travel before it hit the ground?
● Graph this quadratic.

Part 1) The maximum height of the ball is 77.20 feet

Part 2) The maximum height was reached at 2.32 seconds

Part 3) The height of the ball when it was shot was 1.75 feet

Part 4) The ball hit the ground at 4.67 seconds

Part 5) The graph in the attached figure

Step-by-step explanation:

Let

t ----> the time in seconds

h(t) ----> the height of the ball in feet at any time

we have

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

Part 1) Find the maximum height attained by the ball

we know that

The maximum height correspond to the y-coordinate of the vertex

so

Convert the quadratic equation in vertex form

Factor -14

Complete the square

Rewrite as perfect squares

the vertex is the point    

Convert to decimal number

therefore

The maximum height of the ball is 77.20 feet

Part 2) What time was the maximum height achieved?.

we know that

The time when the maximum height was reached is equal to the x-coordinate of the vertex

The vertex is the point  

therefore

The maximum height was reached at 2.32 seconds

Part 3) How high was the ball off of the ground when it was shot?

Find out the y-intercept

The y-intercept is the height of the ball when it was shot (time equal zero)

so

For t=0

therefore

The height of the ball when it was shot was 1.75 feet

Part 4) How far did the ball travel before it hit the ground?

we know that

When the ball hit the ground, the height is equal to zero

so

For h(t)=0

solve the quadratic equation

The formula to solve a quadratic equation of the form

is equal to

in this problem we have

so

substitute in the formula

therefore

The ball hit the ground at 4.67 seconds

Part 5) Graph this quadratic

Plot the given points to graph the quadratic equation

we have

open downward

vertex  (2.32,77.20) ---> is a maximum

y-intercept (0,1.75)

x-intercepts (-0.03,0) and (4.67,0)

The graph in the attached figure

just take the area of the square and then the area of the circle then subtract the circle area from the squares area

step-by-step explanation: