Question 58464
<pre><font size = 5 color = "darkgreen"><b>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 = s<sub>O</sub> + v<sub>O</sub>t + gt<sup>2</sup>/2

Since it starts from the ground, s<sub>O</sub> = 0

Since it ends up on the ground, s = 0

Since its initial speed is 1960 m/s upward, 
v<sub>O</sub> = 1960

Since it is fired on earth, rather than on the 
moon or Mars, g = -9.8 m/s<sup>2</sup>

Plug these all into:

             s = s<sub>O</sub> + v<sub>O</sub>t + gt<sup>2</sup>/2

             0 = 0 + 1960t + (-9.8)t<sup>2</sup>/2

       9.8t<sup>2</sup>/2 = 1960t

Multiply both sides by 2 to clear of fractions

         9.8t<sup>2</sup> = 3920t

 9.8t<sup>2</sup> - 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</pre>