Need to calcultate the answer and round to nearest tenth of a second.

Question

A model rocket is launched upward with an initial velocityof 230 feet per second. The height in feet, of the rocket t seconds after the launch is given by h=-16t +230t. How may seconds after the launch will the rocket be 340 feet above the ground. Round to the nearest tenth of a second.

Answer 1

Since the question gives us the equation that represents the height of the rocket, we can disregard the first sentence of this question.  Since h represents the height above the ground, we can plug the 340 into the equation in place of the h.  Once we have that, we just have one variable to deal with.  We need to find t.  I assume that you meant to put h= -16t²+230t. 

 

Our equation is now 340= -16t²+230t.  The issue with this is that there is a variable that is squared, which means we may need to use the quadratic formula.  We need to subtract the 340 over to the other side, giving us 0= -16t² +230t – 340.  We can make this a bit prettier due to the fact that we have a 0 on one side of the equation and each number on the other side is even.  This means that each number is at least divisible by 2.  If we divide each number by 2, our equation becomes 0= -8t²+115t – 170.  These numbers, unfortunately, do not have a greatest common factor and thus cannot be simplified any more.  With numbers this high, it would most likely be better to use the quadratic formula than attempting to factor this.

 

The quadratic formula is x= (-b±√(b²–4ac))/2a, for a quadratic equation of ax²+bx+c. Our next step is to just plug our numbers into the appropriate spots.  So for all a's, we plug in -8, for all b's we plug in 115 and for all c's we plug in -170.  This gives us x = (-115±√(115²–4(-8)(-170))/2(-8). With all of our multiplying and squaring, we get x = (-115±√(13225–5440))/-16.  This gives us (-115±83.23265)/-16.  Since we have a plus/minus, this means we will have 2 answers.  We will decide which one to use later.  If we add, we get 1.7 seconds. If we subtract, we get 12.7 seconds.

 

Now, we have 2 answers that look like they would work.  How could we have 2 different answers and decide which one to use? Thanks to gravity, what goes up, must come down. (Considering this is a model rocket and not one designed for space travel).  It takes roughly 1.7 seconds to reach the height going up and then it will take roughly 12.7 seconds to be 340 feet high on the way down.  The question is asking how long it will take to reach the height the first time and therefore our answer is 1.7 seconds.

Answer 2

Looking at the basic logic we see the rocket would take more than 1 second and less than 2 seconds to reach a height of 340 feet since it would travel 230 feet in 1 second and 460 feet in 2 seconds.
 
Your initial equation appears to be written incorrectly.  We expect to see a parabolic equation in the form of
h = -at² + bt + c  since the rocket would travel upward initially until it reaches a maximum height then taper off
and fall back to ground.  However, this simplistic equation seems to discount any return to earth and assume the
rocket just continues upward... thus, the equation should be
 
h = 230t + 16   since it would travel 230 feet every second and you need to add the 16ft initial height
 
340 = 230t + 16
230t = 324
t = 324/230
t =  1.409 seconds
 
rounding that answer we get t = 1.4 seconds
 
Check this answer:  230(1.4) + 16 = 338 feet.  Since we rounded this seems correct.

Answer 3

Looking at the basic logic we see the rocket would take more than 1 second and less than 2 seconds to reach a height of 340 feet since it would travel 230 feet in 1 second and 460 feet in 2 seconds.

 

Your initial equation appears to be written incorrectly.  We expect to see a parabolic equation in the form of

h = -at² + bt + c  since the rocket would travel upward initially until it reaches a maximum height then taper off

and fall back to ground.  However, this simplistic equation seems to discount any return to earth and assume the

rocket just continues upward... thus, the equation should be

 

h = 230t + 16   since it would travel 230 feet every second and you need to add the 16ft initial height

 

340 = 230t + 16

230t = 324

t = 324/230

t =  1.409 seconds

 

rounding that answer we get t = 1.4 seconds

 

Check this answer:  230(1.4) + 16 = 338 feet.  Since we rounded this seems correct.

Answer 4

Substitute h=340

 

 

340 = -16t² + 230t

 

 

Subtract 340 on both sides of equation.

 

0 = -16t² + 230t - 340

 

 

Factor out a -2.

 

0 = -2(8t² - 115t + 170)

 

 

Set the term in parenthesis equal to zero.

 

8t² - 115t + 170 = 0

 

 

Use the quadratic formula to find t.

 

t = (115 ± √(13225 - 4(1360)) / 16

 

t = (115 ± √(7785)) / 16

 

t = (115 ± 88.23) / 16

 

t = (115 + 88.23) / 16              and                t = (115 - 88.23) / 16

 

t = 12.7                                  and                 t = 1.7

 

 

1.7 sec is too quick to reach the distance of 340 feet.

 

 

We accept t=12.7

 

It takes 12.7 seconds for the rocket to be 340 feet above the ground.