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using calculus, it can be shown that if a ball is thrown upward with an initial velocity of 1
ft/s from the top of a building 448 ft high, then its height h above the ground t seconds
later will be

during what realistic time will the ball be at least 64 ft above the ground?

minutes

Explanation:

Step1: Set up the inequality

We want \(h\geq64\), so \(448 + 32t-16t^{2}\geq64\).
Rearrange it to \(16t^{2}-32t - 384\leq0\). Divide by 16: \(t^{2}-2t - 24\leq0\).

Step2: Factor the quadratic

Factor \(t^{2}-2t - 24=(t - 6)(t + 4)\).
Set \((t - 6)(t + 4)=0\), we get \(t = 6\) or \(t=-4\).

Step3: Determine the interval

Since the coefficient of \(t^{2}\) in \(y=t^{2}-2t - 24\) (where \(y=(t - 6)(t + 4)\)) is positive (\(a = 1>0\)), the parabola opens upward.
The solution of \(t^{2}-2t - 24\leq0\) is \(-4\leq t\leq6\). But since \(t\geq0\) (time can't be negative in this context), the valid interval is \(0\leq t\leq6\).

Answer:

The ball is at least 64 ft above the ground for \(t\in[0,6]\) seconds. To convert to minutes, since \(1\) minute \( = 60\) seconds, \(t=\frac{6}{60}=0.1\) minutes.