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. 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. height of ball over time

Explanation:

• Initial height: When \(t = 0\), \(h(t)=0\), so the initial height is not 96 feet.
• Ball hitting the ground: When \(h(t) = 0\), \(t = 0\) (initial time) and \(t=5\). So it does not hit the ground between 2 - 3 seconds.
• Maximum height: The function \(h(t)\) is a quadratic function (since the data has a symmetric - like pattern for projectile motion). The formula for the vertex of a quadratic \(y = ax^{2}+bx + c\) is \(t=-\frac{b}{2a}\). For a projectile motion \(h(t)=- 16t^{2}+v_{0}t + h_{0}\). Using the points \((1,64)\) and \((2,96)\):
• Substitute into \(h(t)=-16t^{2}+v_{0}t+h_{0}\), when \(t = 0\), \(h(0)=0\) so \(h_{0}=0\). Then \(h(t)=-16t^{2}+v_{0}t\).
• When \(t = 1\), \(h(1)=-16\times1^{2}+v_{0}\times1=64\), so \(v_{0}=64 + 16=80\). Then \(h(t)=-16t^{2}+80t\).
• The vertex \(t=-\frac{80}{2\times(-16)}=2.5\).
• Also, from the data, the height at \(t = 2\) and \(t = 3\) is 96, but we can calculate \(h(2.5)=-16\times(2.5)^{2}+80\times2.5=-16\times6.25 + 200=-100 + 200 = 100\) (using the formula \(h(t)=-16t^{2}+80t\)), so the maximum is not 96.

Answer:

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