This one reminded me of my days of studying the "three laws of motion".

Well for this one you need neither of those three laws (though this can be solved using them too).

Since we need to find the height after the ball has descended for 7 seconds. We would need the "apex" height that the ball achieves and the time it takes to reach there.

The equation \(h = −3(t − 10)^2 + 250\) could be rewrited as \(h = 250 −3(t − 10)^2 \)

Now, as we need the highest point, \(h_{max}\), it can be maximum only when we subtract nothing from \(250 \) as the quantity \(3(t − 10)^2\) is always going to be positive. So, for that we can put,

\(3(t − 10)^2 = 0\). This gives us, \(t = 10\) seconds. And, \(h_{max}\) = \(250 \) fts.

As the ball has descended for \(7 \) seconds, it will take another \(3 \) seconds to reach the ground, so substituting \(t = 3\) in the given equation will give us the desired height,

\(h_{3secs} = 250 −3(3 − 10)^2 \) => \(h_{3secs} = 103\) ft. (B)
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The maximum height occurs when \(3(t-10)^2\) is as small as possible. That is zero, which is achieved at t = 10. 7 seconds after 10 is 17 seconds. Plug that in and we get:

\(h = −3(17 − 10)^2 + 250\)

\(h = −3(7)^2 + 250\)

\(h = −3(49) + 250\)

\(h = −147 + 250\)

103

Answer choice B.
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