```Question 299015
A ball is thrown vertically upward from the top of a building 102 feet tall with an initial velocity of 60 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s(t)=-16t^2+60t+102
What is the distance of the ball from the ground at time t=0?
Set t=0 and plug into formula:
s(t)=-16t^2+60t+102
s(0)=-16*0^2+60(0)+102
s(0)= 102 feet
.
After how many seconds does the ball strike the ground?
Solve for t when s(t)=-102 (that's because we are 102 feet above the ground)
s(t)=-16t^2+60t+102
-102=-16t^2+60t+102
0=-16t^2+60t+204
0=-16t^2+60t+204
0=-4t^2+15t+51
.
Solve using the quadratic formula.  Doing so yields:
x = {-2.16, 5.91}
Throw away the negative solution leaving:
t = 5.91 seconds
.
After how many seconds will the ball pass the top of the building on its way down?
Solve for t when s(t)=0
s(t)=-16t^2+60t+102
0=-16t^2+60t+102
0=-8t^2+30t+51
Solve using the quadratic formula (Details below).  Doing so yields:
x = {-1.27, 5.02}
Throw away the negative solution leaving:
t = 5.02 seconds
.
Details of quadratic for part c:
*[invoke quadratic "x", -8, 30, 51 ]
```