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