SOLUTION: 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 af

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: 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 af      Log On


   

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?
After how many seconds does the ball strike the ground?
After how many seconds will the ball pass the top of the building on its way down?
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
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:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -8x%5E2%2B30x%2B51+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%2830%29%5E2-4%2A-8%2A51=2532.

Discriminant d=2532 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-30%2B-sqrt%28+2532+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2830%29%2Bsqrt%28+2532+%29%29%2F2%5C-8+=+-1.26993640635228
x%5B2%5D+=+%28-%2830%29-sqrt%28+2532+%29%29%2F2%5C-8+=+5.01993640635228

Quadratic expression -8x%5E2%2B30x%2B51 can be factored:
-8x%5E2%2B30x%2B51+=+-8%28x--1.26993640635228%29%2A%28x-5.01993640635228%29
Again, the answer is: -1.26993640635228, 5.01993640635228. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-8%2Ax%5E2%2B30%2Ax%2B51+%29