Question

1. physics in a physics class demonstration, a ball is dropped from the roof of a building, 72 ft above the ground. the height h, in feet, of the ball above the ground is given by the function h = -16t² + 72, where t is the time in seconds. a. graph the function. b. how far has the ball fallen from time t = 0 to t = 1? c. reasoning does the ball fall the same distance from time t = 1 to t = 2 as it does from t = 0 to t = 1? explain.

Explanation:

Step1: Find initial and final heights at t = 0 and t = 1

When \(t = 0\), \(h(0)=-16(0)^2 + 72=72\) feet. When \(t = 1\), \(h(1)=-16(1)^2+72=-16 + 72 = 56\) feet. The distance fallen from \(t = 0\) to \(t = 1\) is \(72 - 56=16\) feet.

Step2: Find heights at t = 1 and t = 2

When \(t = 2\), \(h(2)=-16(2)^2+72=-64 + 72 = 8\) feet. The distance fallen from \(t = 1\) to \(t = 2\) is \(56 - 8 = 48\) feet.

Step3: Analyze if distances are the same

Since \(16
eq48\), the ball does not fall the same distance from \(t = 0\) to \(t = 1\) as it does from \(t = 1\) to \(t = 2\).

Step4: Graph - general approach

The function \(h(t)=-16t^2 + 72\) is a parabola opening downwards (\(a=-16<0\)) with the vertex at \((0,72)\) (because for \(y = ax^2+bx + c\), the \(x\) - coordinate of the vertex is \(x =-\frac{b}{2a}\), here \(b = 0\)). We can find the \(x\) - intercepts by setting \(h(t)=0\), so \(-16t^2+72 = 0\), \(16t^2=72\), \(t^2=\frac{72}{16}=\frac{9}{2}\), \(t=\pm\frac{3}{\sqrt{2}}\). We consider only \(t\geq0\) for time. Plot the vertex \((0,72)\), the \(x\) - intercept \((\frac{3}{\sqrt{2}},0)\) and a few other points like \((1,56)\) and \((2,8)\) to graph the function.

Answer:

a. To graph \(h(t)=-16t^2 + 72\), plot the vertex \((0,72)\), the \(x\) - intercept \((\frac{3}{\sqrt{2}}\approx2.12,0)\), \((1,56)\) and \((2,8)\) and draw a parabola opening downwards.
b. The ball has fallen \(16\) feet from \(t = 0\) to \(t = 1\).
c. No, the ball does not fall the same distance from \(t = 0\) to \(t = 1\) as it does from \(t = 1\) to \(t = 2\) because it falls \(16\) feet from \(t = 0\) to \(t = 1\) and \(48\) feet from \(t = 1\) to \(t = 2\).