SOLUTION: How long does it take a ball to reach the ground 7.0 m below, if it is thrown straight up with an initial speed of 2.0 m/s?

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Question 892682: How long does it take a ball to reach the ground 7.0 m below, if it is thrown straight up with an initial speed of 2.0 m/s?
Found 2 solutions by Alan3354, Fombitz:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
How long does it take a ball to reach the ground 7.0 m below, if it is thrown straight up with an initial speed of 2.0 m/s?
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9.8 m/sec/sec is commonly used for gravity's acceleration on Earth.
Height as a function of time is:
h(t) = -4.9t^2 + 2t + 7 h in meters, t in seconds
h(t) = 0 at impact
-4.9t^2 + 2t + 7 = 0
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=141.2 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: -1.00844496142963, 1.41660822673575. Here's your graph:

======================
Ignore the negative value
t =~ 1.4166 seconds
Answer by Fombitz(32388)   (Show Source): You can put this solution on YOUR website!



Find t when,

.
.
.

Use the quadratic formula,


Only the positive time makes sense here.

AMP Parsing Error of [sec]: Invalid function '': opening bracket expected at /home/ichudov/project_locations/algebra.com/templates/Algebra/Expression.pm line 70. .
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