# Quadratic Formula Worksheet - Page 3

• Page 3
21.
A stone is dropped from a height of 9 feet above the ground. The height of the stone is modeled by the equation $h$ = - 16$t$2 + 9, where $t$ is the time in seconds. Find the time taken for the stone to hit the ground.
 a. 0.75 b. 1 c. 1.25 d. 0.5

#### Solution:

h = - 16t2 + 9
[Original equation.]

0 = - 16t2 + 9
[Replace h with 0, as the height is zero at the ground level.]

t = {-0±[02-4(-16)(9)]}2(-16)
[Substitute the values in the quadratic formula: a = - 16, b = 0 and c = 9.]

= 0±(0+576)-32
[Simplify.]

0±576-32

= 0±24 / -32

t = -24 / -32 = 0.75
[Since t represents time, use the positive solution.]

22.
Tim runs a textile company that manufactures T-shirts. The profit made by the company is modeled by the function $p$ = $s$2 + 9$s$ - 1620, where $s$ is the number of T-shirts sold. Find the least number of T-shirts to be sold so that Tim does not end up in a loss.
 a. 40 b. 36 c. 42 d. 38

#### Solution:

p = s2 + 9s - 1620
[Original equation.]

0 = s2 + 9s - 1620
[Replace p with 0.]

s = {-(9) ±[9² - 4(1)(-1620)]}2(1)
[Substitute the values in the quadratic formula: a = 1, b = 9 and c = -1620.]

= -9±(81+6480)2
[Simplify.]

= -9±65612

= -9±81 / 2

s = 72 / 2 = 36
[Consider only the positive value of s. s can't be negative as it represents the number of T-shirts.]

23.
A tennis player hits a ball when it is 8 feet off the ground. The ball is hit with an upward velocity of 8 feet per second. After the ball is hit, its height $h$(in feet) is modeled by $h$ = - 16$t$2 + 8$t$ + 8, where $t$ is the time in seconds. How long will it take the ball to reach the ground?
 a. 4 b. 3 c. 2 d. 1

#### Solution:

h = - 16t2 + 8t + 8
[Original equation.]

0 = - 16t2 + 8t + 8
[Replace h with 0, as the height is zero at the ground level.]

t = {-8±[(8)2-4(-16)(8)]}[2(-16)]
[Substitute the values in the quadratic formula: a = - 16, b = 8 and c = 8.]

= -8±(64+512)-32
[Simplify.]

= -8±576-32

= -8±24 / -32

= -32 / -32 = 1
[Since t represents time, discard negative value.]

24.
Chris drops a ball from a height of 64 feet above the ground. Calculate the time taken by the ball to hit the ground, if its height is given by the equation $h$ = - 16$t$2 + 64, where $t$ is the time in seconds.
 a. 1.75 b. 2.5 c. 2.25 d. 2

#### Solution:

h = - 16t2 + 64.
[Original equation.]

0 = - 16t2 + 64
[Substituting 0 for h, since at ground level h = 0.]

t = {-0±[(-0)2-4(-16)(64)]}[2(-16)]
[Substitute the values in the quadratic formula: a = - 16, b = 0 and c = 64.]

= 0±(0+4096)-32
[Simplify.]

= 0±4096-32
[Simplify inside the grouping.]

= 0±64 / -32

= -64 / -32 = 2
[Since t represents time, use the positive solution.]

So, the ball takes 2 sec to reach the ground.

25.
Sunny throws a pencil from a building with an initial downward velocity of - 8 feet per second. How long will the pencil take to reach the ground, if the height of the pencil from the ground is modeled by the equation $h$ = - 16$t$2 - 8$t$ + 48, where $t$ is the time in seconds?
 a. 2 b. 1.5 c. 1.75 d. 1.25

#### Solution:

h = - 16t2 - 8t + 48
[Original equation.]

0 = - 16t2 - 8t + 48
[Substitute values and write in the standard form.]

t = {-(-8)±[(-8)2-4(-16)(48)]}2(-16)
[Substitute the values in the quadratic formula.]

= 8±(64+3072)-32
[Simplify.]

= 8±3136-32

= 8±56 / -32
[Simplify.]

= -48 / -32 = 1.5
[Evaluating the radical and taking the positive value since t represents time.]

26.
What are the values of $a$, $b$ and $c$ in the equation 2$f$ 2 - 8$f$ + 22 = 0, which is in the standard form?
 a. $a$ = 2, $b$ = - 8 and $c$ = 22 b. $a$ = -2, $b$ = 8 and $c$ = - 22 c. $a$ = 2, $b$ = 22 and $c$ = 0 d. $a$ = - 2, $b$ = - 8 and $c$ = 22

#### Solution:

The standard quadratic equation is ax2 + bx + c = 0 where a ≠ 0.

2f 2 - 8f + 22 = 0
[Original equation.]

a = 2, b = - 8 and c = 22
[Compare the original equation with the standard equation.]

27.
Write the equation which is used to model the height of an object that is thrown down with an initial velocity of - 8 feet per second from a height of 18 feet.
 a. $h$ = - 16$t$2 + 8$t$ + 18 b. $h$ = - 16$t$2 + 18 c. $h$ = - 16$t$2 - 8$t$ + 18 d. $h$ = - 16$t$2 - 8$t$ - 18

#### Solution:

h = - 16t2 + vt + s
[Model for an object which is thrown from an initial height 's', with an initial velocity 'v' and the time in motion 't'.]

h = - 16t2 - 8t + 18
[Substitute v = - 8 and s = 18.]

Therefore, the equation of an object that is thrown down with an initial velocity of - 8 feet per second from a height of 18 feet is h = - 16t2 - 8t + 18.