Question

lara made the table below of the predicted values for ( h(t) ), the height, in meters, of a penny ( t ) seconds after it is dropped off of the back of the bleachers. to the nearest tenth of a second, how much time would it take the penny to hit the ground?

Explanation:

Step1: Analyze the height values

We are looking for the time \(t\) when \(h(t)\) changes from positive (above ground) to negative (below ground, which means it has hit the ground).

Step2: Check the table values

At \(t = 0.6\), \(h(t)=0.236\) (positive, penny is still in the air). At \(t = 0.7\), \(h(t)=- 0.401\) (negative, penny has hit the ground). Since we want the time when it just hits the ground (crosses \(h(t) = 0\)), and we are to the nearest tenth of a second, we note that the change from positive to negative occurs between \(t = 0.6\) and \(t=0.7\). But if we assume a linear - like behavior (for the purpose of choosing from the given options) and check the options:

• Option 0.5: \(h(0.5)=0.775\) (penny is still above ground)
• Option 0.6: \(h(0.6) = 0.236\) (penny is still above ground)
• Option 0.7: \(h(0.7)=-0.401\) (penny has passed the ground level)
• Option 0.8: \(h(0.8)=-1.136\) (penny is well below ground level, but we want the earliest time when it hits the ground)

Since the penny hits the ground when \(h(t)\) crosses \(0\) and we consider the nearest - tenth - of - a - second approximation from the table values (where at \(t = 0.7\) it is the first time \(h(t)\) is negative), we choose \(t = 0.7\)

Answer:

0.7 seconds