Question

a ball is thrown straight up into the air. the table shows the data collected over t seconds, where h(t) is the height of the ball, in feet. height of ball over time which statement is true? the initial height of the ball is 96 feet. the ball will hit the ground between 2 and 3 seconds after it was thrown. the maximum height of the ball must be 96 feet. the maximum height of the ball was reached 2.5 seconds after it was thrown.

Explanation:

Step1: Analyze the initial height

When \(t = 0\), \(h(0)=0\). So the initial height is \(0\) feet, not \(96\) feet.

Step2: Analyze when the ball hits the ground

The ball hits the ground when \(h(t)=0\). At \(t = 0\) (initial throw) and \(t = 5\) \(h(t)=0\). So it does not hit the ground between \(2\) and \(3\) seconds.

Step3: Analyze the maximum height

The function \(h(t)\) is a quadratic function (since the height - time relationship for a projectile is quadratic \(h(t)=at^{2}+bt + c\)). For a quadratic function \(y = ax^{2}+bx + c\), the vertex \(t=-\frac{b}{2a}\). Also, for a quadratic \(h(t)\) passing through \((1,64)\), \((2,96)\), \((3,96)\), \((4,64)\). The maximum of a quadratic \(y = ax^{2}+bx + c\) occurs at \(t=\frac{2 + 3}{2}=2.5\) (because for a quadratic \(y = ax^{2}+bx + c\) symmetric about \(x =-\frac{b}{2a}\), and since \(h(2)=h(3) = 96\), the axis of symmetry is \(t=\frac{2+3}{2}=2.5\)). The maximum height is \(h(2.5)\). We can find the quadratic function. Let \(h(t)=at^{2}+bt + c\). Using \((0,0)\): \(c = 0\). Using \((1,64)\): \(a + b=64\). Using \((2,96)\): \(4a+2b=96\) or \(2a + b = 48\). Subtract \(a + b=64\) from \(2a + b=48\): \((2a + b)-(a + b)=48 - 64\), \(a=-16\), then \(b = 80\). So \(h(t)=-16t^{2}+80t\), and \(h(2.5)=-16\times(2.5)^{2}+80\times2.5=-16\times6.25 + 200=-100 + 200 = 100\) (but the key is the time of maximum).

Answer:

The maximum height of the ball was reached \(2.5\) seconds after it was thrown.