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
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)
1 answer
3 answers