Question

If an object is propelled upward from a height of s feet at an initial velocity of v feet per seconds, then its height h after t seconds is given by the equation h=-16t^2+vt+s where h is in feet. If the object is propelled from a height of 12 feet with an initial velocity of 64 feet per seconds its height h is given by the equation h=-16t^2+64t+8
After how many seconds is the height 40 feet?

Answer 1

68=-16t2+64t+8
16t2-64t+60=0
4t2-16t+15=0
(2t-5)(2t-3)=0
t=5/2 or 3/2
The object is at 68 feet after 3/2 secs, and returns to 68 feet after another second

Answer 2

Same question as last one but with different numbers

Answer 3

oh ok