Question 970359
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.