Question

If an object is propelled upward from a height of s feet at an initial velocity of v feet per​ second, then its height h after t seconds is given by the equation h=−16 t squared +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​ second, its height h is given by the equation h=-16t squared+64t+12. After how many seconds is the height 72 ​feet?

Answer 1

From the given information we can write the equation:

-16t^2 +96t + 4 = 112

-16t^2 + 96t + 4 - 112 = 0

-16t^2 + 96t - 108 = 0

simplify, change the signs, divide by -4

4t^2 - 24t + 27 = 0

You can use the qudratic formula but this will factor to

(2t-3)(2t-9) = 0

Two solutions

t = 3/2

t = 1.5 seconds at 112 ft on the way up

and

t = 9/2

t = 4.5 seconds at 112 ft on the way back down

Graphically, (green line is 112 ft)