26 January, 08:32
If a ball is thrown directly upward with a velocity of 50ft/s, it's height (in feet) after t seconds is given by y=50t-16t^2 what is the maximum height attained by the ball
26 January, 09:03
Since the function is a quadratic representing height, and the coefficient of the t² is negative, the vertex of the parabola will be the maximum height achieved by the ball.
The general form for a quadratic equation is ax² + bx + c, here a is - 16, and b is 50
To find the x coordinate of the vertex, use x = - b / (2a)
We have x = - 50/[2 (-16) ]
x = - 50/-32
x = 25/16
Now plug that into the equation to find the y value, which will be the height ...
y = 50 (25/16) - 16 (25/16) ²
y = 1250/16 - 16 (625/256)
y = 1250/16 - 625/16
y = 625/16
y = 39.0625
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» If a ball is thrown directly upward with a velocity of 50ft/s, it's height (in feet) after t seconds is given by y=50t-16t^2 what is the maximum height attained by the ball