SOLUTION: The path of a baseball thrown in an arch the shape of a parabola is given by the quadratic equation y= -16t^2+2t+10, where T is time in seconds and Y is the height of the ball in i
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-> SOLUTION: The path of a baseball thrown in an arch the shape of a parabola is given by the quadratic equation y= -16t^2+2t+10, where T is time in seconds and Y is the height of the ball in i
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Question 436825: The path of a baseball thrown in an arch the shape of a parabola is given by the quadratic equation y= -16t^2+2t+10, where T is time in seconds and Y is the height of the ball in its flight through the air.
Use the quadratic Formula to find at what time,T, the ball hits the ground. Write your answer in exact form and as a decimal rounded to the nearest hundredth. [Hint: y=0 when the ball hits the ground.] Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! There something seriously wrong with the equation. 2t represents the upward velocity of the ball.
That is not even enough to overcome gravity (-16t^2)
10 represent the initial height of the ball when it is thrown, possible but not likely either
