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Re: The height, h, of a thrown ball as a function of t, the amount of time [#permalink]
1
Hey,

This is an easy one but only if you know differentiation. Let's try the other way.

Consider it as a quadratic equation. h(t)=\(–10t^2+40t\)

It's an equation of parabolic curve. 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 \(-10\), and \(b\) is \(40\)
Now maximum height will occur at \(t = \frac{-b}{2a}\)

\(t = \frac{-40}{-20} = 2\)

Put the value in the original equation,

h(2)=\(–10(2)^2+40(2)\) = 40

Maximum height = 40

Answer C

Please refer this for any doubts : https://brainly.com/question/12298041


arjunbir wrote:
I still did not get the explanation. Can someone please explain the solution in detail?
thanks
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Re: The height, h, of a thrown ball as a function of t, the amount of time [#permalink]
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