Mathematics, 26.03.2020 16:17, Dreamer05

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.

Answers: 1

Show answers

Answers

answered: hehefjf8610

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

h(t)=-14t^2+65t+1.75

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

Convert the quadratic equation in vertex form

h(t)=-14t^2+65t+1.75

Factor -14

$h(t)=-14(t^2-\frac{65}{14}t)+1.75$

Complete the square

$h(t)=-14(t^2-\frac{65}{14}t+\frac{4,225}{784})+1.75+\frac{4,225}{56}$

$h(t)=-14(t^2-\frac{65}{14}t+\frac{4,225}{784})+\frac{4,323}{56}$

Rewrite as perfect squares

$h(t)=-14(t-\frac{65}{28})^2+\frac{4,323}{56}$

the vertex is the point $(\frac{65}{28},\frac{4,323}{56})$

Convert to decimal number

(2.32,77.20)

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 (2.32,77.20)

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)

For t=0

$h(t)=-14(0)^2+65(0)+1.75=1.75\ ft$

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

For h(t)=0

-14t^2+65t+1.75=0

solve the quadratic equation

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

-14t^2+65t+1.75=0

$a=-14\\b=65\\c=1.75$

substitute in the formula

$t=\frac{-65\pm\sqrt{65^{2}-4(-14)(1.75)}} {2(-14)}$

$t=\frac{-65\pm\sqrt{4,323}} {-28}$

$t=\frac{-65+\sqrt{4,323}} {-28}=-0.03$

$t=\frac{-65-\sqrt{4,323}} {-28}=4.67$

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

A ball is shot from a cannon into the air with an upward velocity of 65 ft/sec. The equation that gi

answered: Guest

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

u also answered this so i think u would know but i am here to

put in or use mine