Graham and hunter are circus performers. a cable lifts graham into the air at a constant speed of 1.5 ft/s. when graham’s arms are 18 ft above the ground, hunter, who is standing directly underneath graham, throws graham a ball as the cable continues to lift him higher. hunter throws the ball from a position 5 ft above the ground with an initial velocity of 24 ft/s. which system of equations can be used to model this situation?
a) h=18+1.5t
h=5+24t-16t^2
b) h=18+1.5t
h=5+24t+16t^2
c) h=18+1.5t
h=5+24t
d) h=18t+1.5t-16t^2
h=5+24t-16t^2

Hello there!

this type of equations in one dimension (when all the factors are constants) are written as:

h =  initial position +  initial velocity*t + (acceleration/2)*t^2

First, let's describe the hunter's equation:

We know that Graham moves with a velocity of 1.5 ft/s, and when he is  18 ft above the ground, Hunter throws the ball, and because Graham is pulled with a cable, he is not affected by gravity.

If we define t= 0 when Graham is 18 ft above the ground, the equation for Graham height (in feet) is:

h = 18 + 1.5t

where t in seconds.

Now, the equation for the ball:

We know that at t= 0, the ball is thrown from an initial distance of 5ft, with an initial velocity of 24ft/s and is affected by gravity acceleration g, where g is equal to: 32.2 ft/s (notice that the gravity pulls the ball downwards, so it will have a negative sign)

the equation for the ball is:

h = 5 + 24t - (32.2/2)t^2 = 5 + 24t - 16.1t^2

So the system is:

h = 18 + 1.5t

h = 5 +24t - 16.1t^2

so the right answer is A

-2

7 - 9 = -2

reversed:

-2 + 9 = 7

the answer is y < -3 or   y > 3

-3y > 7 + 2

-3 y > 9

y < -3