SOLUTION: A baseball player hits a baseball into the air from a height of 3 feet with an initial vertical velocity of 72 feet per second. After how many seconds is the baseball 84 feet above
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Question 189928This question is from textbook McDougal Littell Math
: A baseball player hits a baseball into the air from a height of 3 feet with an initial vertical velocity of 72 feet per second. After how many seconds is the baseball 84 feet above the ground?
This question is from textbook McDougal Littell Math
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A baseball player hits a baseball into the air from a height of 3 feet with
an initial vertical velocity of 72 feet per second. After how many seconds
is the baseball 84 feet above the ground?
:
The equation for this is:
h = -16t^2 + 72t + 3
Where:
h = height at t seconds
-16t^2 is the downward pull of gravity
72t is the vertical velocity upward
3 is the initial height at t=0
:
Find t when height is 84 ft
-16t^2 + 72t + 3 = 84
Arrange as a quadratic equation:
-16t^2 + 72t + 3 - 84 = 0
-16t^2 + 72t - 81 = 0
multiply equation by -1 to change the signs, easier to factor
16t^2 - 72t + 81 = 0
this is a perfect square
(4t - 9)(4t - 9) = 0
4t = +9
t =
t = 2.25 sec it will be at 84' (this is also the highest point (vertex))
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