Question
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
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