Asked by Alex
A projectile is shot vertically into the air. Its height, h, in metres, after t seconds is approximately modelled by the relation h=-5t^2+120t+125:
c) For how many seconds is the projectile in the air? (1 mark)
c) For how many seconds is the projectile in the air? (1 mark)
Answers
Answered by
Reiny
According to your equation, the object was 125 m above ground (t=0) , when it was shot
solve for h = 0 , this will give you the two times it is on the ground
-5t^2 + 120t + 125 = 0
divide by -5
t^2 - 24t - 25 = 0
(t-25)(t+1) = 0
t = -1 or t = 25
but, this only makes sense for t≥ 0, so the object was in the air from 0 to 25 seconds
or 25 seconds
solve for h = 0 , this will give you the two times it is on the ground
-5t^2 + 120t + 125 = 0
divide by -5
t^2 - 24t - 25 = 0
(t-25)(t+1) = 0
t = -1 or t = 25
but, this only makes sense for t≥ 0, so the object was in the air from 0 to 25 seconds
or 25 seconds
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.