SOLUTION: a rocket is fired upward with an inital speed of 1960 m/s. after how many minutes does it hit the ground?

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Question 58464This question is from textbook Algebra structure and Method
: a rocket is fired upward with an inital speed of 1960 m/s. after how many minutes does it hit the ground? This question is from textbook Algebra structure and Method

Found 2 solutions by 6894, Edwin McCravy:
Answer by 6894(1)   (Show Source): You can put this solution on YOUR website!
let t be the time when the rocket hits the ground.
0=1960t-4.9t^2
0=4.9t(400-t)
4.9t=0, t=0 (when the rocket hits the ground)
400-t=0
400=t
the rocket hits the ground after 400 seconds or 6 2/3 min.

formula used: h=rt-4.9t^2
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
a rocket is fired upward with an inital speed of 
1960 m/s. after how many minutes does it hit the 
ground?


The formula is

             s = sO + vOt + gt2/2

Since it starts from the ground, sO = 0

Since it ends up on the ground, s = 0

Since its initial speed is 1960 m/s upward, 
vO = 1960

Since it is fired on earth, rather than on the 
moon or Mars, g = -9.8 m/s2

Plug these all into:

             s = sO + vOt + gt2/2

             0 = 0 + 1960t + (-9.8)t2/2

       9.8t2/2 = 1960t

Multiply both sides by 2 to clear of fractions

         9.8t2 = 3920t

 9.8t2 - 3920t = 0

t(9.8t - 3920) = 0

Setting the first factor = 0

             t = 0

Setting the second factor = 0

   9.8t - 3920 = 0
 
          9.8t = 3920

             t = 3920/9.8

             t = 400 seconds or 6 2/3 minutes, or 

6 minutes 40 seconds.

So the rocket is on the ground at time 0 
(at lift-off), which is why we get t = 0 as one 
solution. Then the rocket is again on the ground 
6 2/3 minutes later, which is the solution we
are looking for.

Edwin
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