SOLUTION: Hi can you please help me with this question. A golf ball is hit from the top of a tee. The quadratic equation y=-5x^2 + 20x + 0.05 describes its height,y, in metres as time,x, in

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Question 1183267: Hi can you please help me with this question. A golf ball is hit from the top of a tee. The quadratic equation y=-5x^2 + 20x + 0.05 describes its height,y, in metres as time,x, in seconds passes. Determine how long the ball is in the air. Thanks for the help
Found 2 solutions by MathLover1, MathTherapy:
Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!

A golf ball is hit from the top of a tee. The quadratic equation

describes its height, , in metres as time,, in seconds passes.
Determine how long the ball is in the air.
First, you have to determine how long it will take for the golf ball to reach its maximum height.
so, rewrite equation in vertex form
where and are coordinates of vertex
....group
....factor out

....



=> and
to reach its maximum height of , it will take seconds
The time in the air is twice the one I found solving for t where the height is maximum.
the ball is seconds in the air


Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

Hi can you please help me with this question. A golf ball is hit from the top of a tee. The quadratic equation y=-5x^2 + 20x + 0.05 describes its height,y, in metres as time,x, in seconds passes. Determine how long the ball is in the air. Thanks for the help
Since time is x, the ball will be on the ground when y, or height = 0

Therefore, all you need to do is solve the quadratic. 

Using the quadratic equation formula, we get 2 solutions, as expected. They are: - .0025 and 4.0025 seconds.

Obviously, time, in this case, CANNOT be < 0 (negative), so time the ball was in the air is the LATTER. 

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