A few days ago
Height of a Ball? College Algebra Math?
I need help on how to set up this problem. If a ball is thrown directly upward with a velocity of 40 ft. per sec., its height (in feet) after t seconds is given by y=40t-16t squared. What is the max height attained by the ball??
Top 3 Answers
A few days ago
Favorite Answer
f(x) = -16t² + 40t is the equation in question. Since it is a quadratic, we can find the Axis of symmetry by calculation. the line is found where t = -b/(2a) . this gives you t = (5/4) so the maximum height is when the time is 1.25seconds. plug that in place of t and get a maximum height of 45ft. I did this off the top of my head so you might want to double check.
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A few days ago
It’s position at initial and final point is 0
0=40t-16t²
0=t(40-16t)
t=0,5/2
since the equation describes a parabola the vertex(maximum) should occur halfway between the zeros or at t=5/4
y=40(5/4)-16(5/4)²
=50-25
=25 ft.
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A few days ago
40/32 ft is maximum time
plug that in to find max height
y = -16t^2 + 40t is the equation
take the derivative of that and equal it to 0 to find when the function isnt changing anymore (maximum or minimum of a function)
y’ = -32t + 40
0 = -32t + 40
32t = 40
t = 40/32
y = -16t^2 +40t (original function, now plug in T to find max height)
y = -16(40/32)^2 + 40(40/32) is the answer, and I have no intention of calculating the exact number 🙂
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