Cheryl and daryl are circus performers. a cable lifts cheryl into the air at a constant speed of 2 ft/s. when cheryl's arms are 12 ft above the ground, daryl throws her a ball. he throws the ball from a height of 6 ft with an initial vertical velocity of 24 ft/s.

The statement is incomplete.

This is the full statement, with the question:

Cheryl and Daryl are circus performers. A cable lifts Cheryl into the air at a constant speed of 2 ft/s. When Cheryl's arms are 12 ft above the ground, Daryl throws her a ball. He throws the ball from a height of 6 ft with an initial vertical velocity of 24 ft/s.

Identify Variables

t = time, in seconds, since the ball was thrown

h = height, in feet, above the ground

Cheryl's height = initial height + change in height

h=12+2t

Projectile Motion Formula: −16t2+vt+h0

How to Interpret the Solutions to the System

h = 12+2t

h= − 16t² +24t + 6

i) the solution of the system is when both ball and arms are at the same height.

ii) that happens two times, first time at 0.375 s of the ball have been launched, at the height of 12.75 ft, and second time 0.1 s after have been launched at the heigh of 14 ft.

Explanation:

1) Identify the question: you just need to explain how the two equations represent the situation and find the solution.

2) The first equation h = 12 + 2t is the heigth of the arms of Cheryl, where:

i) 12 is the height at which they were at the instant the ball was thrown to her and

ii) 2t is the product of the speed of the cable that pulled her (2 ft/s) by the time.

3) The second equation h= − 16t² +24t + 6 is the height of the ball, where:

i) the first term ( -16 t²) is half the gravitational acceleration (-32 ft/s²) times the time squared,

ii) the second term (24t) is the product of the velocity with which the ball was thrown (initial velocity) times the time, and

iii) the third term (6) is the height from which the ball was thrown.

2) Then, the solution of the system is the moment and heigth at whic the ball thrown vertically upward meet the arms of Cheryl: both ball and arms are at the same height.

3) The solution is found by solving the system:

h = h ⇒ 12+2t = − 16t² +24t + 6

These are the steps:

i) transpose all the terms to one side:

− 16t² + 24t + 6 - 12 - 2t = 0

ii) combine like terms:

− 16t² + 22t - 6 = 0

iii) divide by - 2:

8t² - 11 t + 3 = 0

iii) complete squares and factor the perfect square trinomial:

8t² - 11t = - 3

8 (t² - 11/8 t) = - 3

8(t² - 11/8t + 121/256) = - 3 + 8(121/256)

8(t - 11/16)² = - 3 + 121/32

iv) solve for t:

8(t - 11/16)² = 25/32

(t - 11/16)² = 25/256

t - 11/16 = (+/-) 5/16

t = 11/16 (+/-) 5/16

t = 16/16 = 1 and t = 6/16 = 0.375

Then, the ball meets the arms of Cheryl first time at 0.375 s of have been launched.

And the height was 12 + 2(0.375) = 12.75 ft

The ball meets the arms of Cheryl second time at 0.1 s after have been launched and the heigh was 12 + 2(1) = 14 ft.

If you have the co-ordinate pairs (x, y), (r, s), and (a, b) for the triangle, perform the operations (3x, 3y), (3r, 3s), (3a, 3b). These will give you the knew points since the scale factor has been applied.