Question

A child lets go of a balloon that rises at a constant rate. 5 seconds after it was released, the balloon is at a height of 16 feet above the ground. After 20 seconds of rising, it is at a height of 52 feet above the ground. 1. Write a linear model for the height, h, of the balloon as a function of the number of seconds, s that it has been raising. 2. What was the height of the balloon initially before the child let it go? 3. Use your model to predict the height of the balloon after 90 seconds. Only an algebraic solution will receive credit.

Answer 1

Answer:

1. h = 2.4t + 4

2. 4 feet

3. 220 feet

Step-by-step explanation:

1. Write a linear model for the height, h, of the balloon as a function of the number of seconds, s that it has been raising.

Since the balloon rises at a constant rate, we find this rate by using the initial and final values of height and time at 5 seconds and 20 seconds respectively which are 12 feet and 52 feet respectively.

So, rate = gradient of line

= change in height/change  in time

= (52 ft - 16 ft)/(20 s - 5 s)

= 36 ft/15 s

= 2.4 ft/s

Now the equation of the line which shows the height is gotten from

(h - h')/(t - t') = rate

Using h'= 16 feet and t' = 5 s, we have

(h - 16)/(t - 5) = 2.4

h - 16 = 2.4(t - 5)

h - 16 = 2.4t - 12

h = 2.4t - 12 + 16

h = 2.4t + 4

where h is the height of the balloon above the ground and t is the time spent in the air in seconds.

2. What was the height of the balloon initially before the child let it go?

We obtain the initial height of the balloon before the child let go at time, t = 0 the time before the child let go.

So, substituting t = 0 into the equation for h, we have

h = 2.4t + 4

h = 2.4(0) + 4

h = 0 + 4

h = 4 feet

So, the height of the balloon before the child let go is 4 feet above the ground.

3. Use your model to predict the height of the balloon after 90 seconds.

We insert t = 90 s into the equation for h. So,

h = 2.4t + 4

h = 2.4(90) + 4

h = 216 + 4

h = 220 feet

So, the height of the balloon after 90 s is 220 feet above the ground.