Algebra 1, Please Help! Describing Situations

Question

The height, h, of a stomp rocket (propelled by a short blast of air) above the ground after t seconds is given by the equation h(t) = 5 + 100t − 16t².

Here is a graph that represents h.

https://assets.learnosity.com/organisations/154/40b2c09b-4ca1-4947-a1a7-f69ca3109581.PNG 

In your answer space, number each answer in order to keep organized with your work.

 

1. How does the 5 in the equation relate to the graph? How do you know?

 

2. What does 100t in the equation mean in terms of the rocket? How do you know?

 

3. What does the −16t² mean in terms of the rocket? How do you know?

 

4. About when does the rocket hit the ground? (Give a good estimate) How do you know?

Answer 1

5 is the y or h intercept (0,5)

general quadratic equation is y=(a/2)t^2 +vot + ho, where ho = the initial height, = the constant term

100 = vo = the initial vertical velocity

100t = the iniital velocity times the number of seconds the rocket is in the air

-16t^2 means a= -32 which is the effect of gravity = -32 feet per second per second = the rate of acceleration, negative acceleration or deceleration as it slows down going up, but speeds up going down

it hits the ground when h(t) = 0 = -16t^2 + 100t +5

zero height is ground level

solve for t, using the quadratic formula: t=-b/2a + or - (1/2a)sqr(b^2-4ac) b=100, a=-16, c=5

t =-100/-32 + or - (1/32)sqr(100^2 +4(16)(5))

t = 3 1/8 + or - (1/32)(sqr(10320)

t = 3.125 + or - (1/32)(101.587)

t =3.125 + or - 3.175

t= 6.3 seconds to hit the ground (ignore the negative solution)

A little over 6 1/4 seconds

it takes 3 1/8 seconds to reach maximum height. It takes slightly more than twice that. Had the original height been zero, it would have been exactly 2(3 1/8) = 6 1/4. But it takes another part of second to go from 5 feet high to zero height. 6.25 + But call it a little over 6 seconds either way, to hit the ground.

max height is when h'(t) = 0, when -32t + 100=0, when t = 100/32 = 3.125 seconds

twice that is 6.25 seconds add a little more for that last 5 feet fall to the ground.

6.3 seconds to hit the ground

Answer 2

1. The "5" represents the initial height of the rocket (the height it was before it was launched). You know because you can plug in t = 0 into the equation and you get h = 54
2. Poorly worded question. The "100", which is the coefficient in front of the "t" is the initial velocity in ft/s. We know that because it is the coefficient in front of the "t" but more specifically, if you plug in a small time, (like t = 0.01), you will see that the rocket is traveling at that speed. (at t = 0.01, the height is 5.9984 so the rocket has traveled 0.9984 feet in 0.01 s and 0.9984/0.1 = 99.84 ft/s. If you plug in t = 0.001, the height is 5.099984 meaning the rocket traveled 0.099984 ft in 0.001 s = 99.984 ft/s). The closer you get to t = 0, the closer you get to the initial velocity of 100 ft/s
3. Again, a poorly worded question but the "-16" represents the acceleration due to gravity, which is actually -32 ft/s² but the math works out that this is divided by 2 in the equation. We know this represents the acceleration because it is the coefficient in front of the t² term. It's a little harder to see this in terms of the graph but if you are really looking for the answer, look at this table:

time height 1st difference 2nd difference

0 5

84

1 89 -32

52

2 141 -32

20

3 161 -32

-12

4 149 -32

-44

5 105 -32

-76

6 29 -32

Notice that the second difference is always -32 which is the acceleration due to gravity.

1. The rocket hits the ground at what looks to me as 6.3 seconds