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a) b) We have to start by labeling the graph.
This graph relates the height in the vertical axis with the distance in the horizontal axis. The equation that relates y and x is different from h(t), as we are not representing time in the horizontal axis.
Then, both the height and the distance will have units of feet:
The highest point will be at the point where the height stop increasing and start decreasing.
c) We can use the equation fo h(t) to find the value of t when h(t) = 0, that is , when the ball touches the ground.
As h(t) is a quadratic equation, finding t for h(t) = 0 is finding the roots of the quadratic equation:
[tex]\begin{gathered} h(t)=-16t^2+29t+6 \\ t=\frac{-29\pm\sqrt[]{29^2-4\cdot(-16)\cdot6}}{2\cdot(-16)} \\ t=\frac{-29\pm\sqrt[]{841+384}}{-32} \\ t=\frac{-29\pm\sqrt[]{1225}}{-32} \\ t=\frac{29\pm35}{32} \\ t_1=\frac{29-35}{32}=-\frac{6}{32}=-0.1875 \\ t_2=\frac{29+35}{32}=\frac{64}{32}=2 \end{gathered}[/tex]As the first root is a negative number, it does not make sense in this case. The solution then is the other root, that has a value of t=2. As t is in seconds, we know that the ball reaches the ground 2 seconds after the launch.
Answer:
a) The labels and units are Height (in feet) for the vertical axis and Distance (in feet) for the horizontal axis.
b) The highest point corresponds to the point where the height stops increasing and starts decreasing.
c) The ball touches the ground 2 seconds after the launch.
