Question 1076898
If the object is launched vertically at an initial velocity of 192 ft./sec, then the equation describing it's motion would be -16tē+192t, where t is time, in seconds, after launch.
a)The time needed to reach maximum altitude is given by the expression -b/2a, for an equation of the form atē+bt+c. In this case we get the value of:
-b/2a=-192/-32=6
After 6 seconds our object is at a height of h=-16(6ē)+192(6), or 576 ft.
b)When the object hits the ground, its' height is 0, so:
0=-16tē+192t
16tē=192t
16t=192
t=12
The object hits the ground after 12 seconds. ☺☺☺☺