SOLUTION: Hugh Betcha launched a model rocket with an initial speed of 88 feet per second. After how many seconds will the rocket be 40 feet high? Note: The approximate height h (in feet) is

Question 970359: Hugh Betcha launched a model rocket with an initial speed of 88 feet per second. After how many seconds will the rocket be 40 feet high? Note: The approximate height h (in feet) is modeled by: h=-16t^2+vt, where t is the time in motion (in seconds) and v is the initial upward velocity (in feet per second).
Answer by amarjeeth123(569) (Show Source): You can put this solution on YOUR website!
Substituting the initial velocity we get,
h=-16t^2+88t
For 40 feet we have,
40=-16t^2+88t
16t^2-88t+40=0
2t^2-11t+5=0
2t^2-10t-t+5=0
2t(t-5)-1(t-5)=0
(2t-1)(t-5)=0
t=5 sec.
At t=5 sec the rocket is at 40 feet from the ground.
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