SOLUTION: A ball is thrown off the top of a classroom building with an initial upward velocity of 10 feet per second. The height above the ground can be found using h= -16t(squared) + 10t +

Question 613244: A ball is thrown off the top of a classroom building with an initial upward velocity of 10 feet per second. The height above the ground can be found using h= -16t(squared) + 10t + 40 with t representing time in seconds and h representing height in feet. How high is the ball after 0.8 seconds? When will the ball be 20 feet above the ground and headed down?
Found 2 solutions by Alan3354, nerdybill:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A ball is thrown off the top of a classroom building with an initial upward velocity of 10 feet per second. The height above the ground can be found using h= -16t(squared) + 10t + 40 with t representing time in seconds and h representing height in feet. How high is the ball after 0.8 seconds?
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Sub 0.8 for t
h(0.8) = -16*0.8^2 + 10*0.8 + 40
= 37.76 feet
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When will the ball be 20 feet above the ground and headed down?
h = 20
20 = -16t^2 + 10t + 40

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=1380 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: -0.848385976312919, 1.47338597631292. Here's your graph:

=======================
Ignore the negative solution.

Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
A ball is thrown off the top of a classroom building with an initial upward velocity of 10 feet per second. The height above the ground can be found using h= -16t(squared) + 10t + 40 with t representing time in seconds and h representing height in feet. How high is the ball after 0.8 seconds?
.
given:

set t to 0.8 and solve for h:

feet
.
When will the ball be 20 feet above the ground and headed down?
set h to 20 and solve for t:

Applying the "quadratic formula" we get:
x = {-0.46, 1.09}
answer:
x = 1.09 secs
.
details of quadratic follows:

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=153 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 1.08558230480331, -0.460582304803311. Here's your graph:

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