Word Problem

Question

NASA launches a rocket at t=0 seconds. Its height, in meters above sea-level, in terms of time is given by h(t)=−4.9t^2+106t+388. For each of these questions, report your answer to the nearest tenth.
How high is the rocket after 9 seconds?


How high was the rocket when it was initially launched?



When will the rocket reach its maximum height?



What will its maximum height be (above sea level)?



If the rocket is launched so that it lands in the ocean, when will it splash down?

Answer 1

Hi Kelani.

 

Your formula of interest is:

 

h(t)= −4.9t^2 + 106t + 388

 

a) Initially launched is a code term for "t=0". Plug in 0 for t and determine h(t)

 

h(0) = -4.9(0)^2 + 106(0) + 388

h(0) = 388

 

b) The rocket will reach maximum height based on a formula you may or may not remember from Algebra 1 concerning quadratic equations:

 

t_max = -b/2a

 

In your case, b = 106, a = -4.9

 

So

 

t_max = -106/-9.8

t_max = 10.82 seconds

 

The other way to determine the max height is to take the derivative of h(t) and set it equal to 0.

 

h'(t) = -9.8t + 106

h'(t) = 0

0 = -9.8t + 106

9.8t = 106

t = 106/9.8

t = 10.82 seconds

 

So you yield the same result.

 

c) The maximum height will occur at the maximum time. So you want to plug in t_max for h(t)

 

h(t_max) = -4.9*(10.82)^2 + 106(10.82) + 388

 

h(t_max) = 961.7 ft

 

d) Determining when it splashes is the same thing as asking when the height of the rocket is 0. So set h(t) equal to 0 and determine the t's which make it happen.

 

0 = −4.9t^2 + 106t + 388

 

Here, since you have a quadratic equation, you can employ the quadratic formula to solve for t.

 

t = (-b +/- √(b^2 - 4ac))/2a

 

a = -4.9

b = 106

c = 388

 

t = -3.2, 24.8

 

Time can't be negative so we can throw the first term out.

 

Your rocket will splash down in 24.8 seconds.

 

Hope this helps.