Respuesta :
We have the expression for the height H of the rock in function of the time r, when it falls from a height of 20 meters:
[tex]H(r)=20-4.9\cdot r^2[/tex]When r = 1 second, we have:
[tex]H(1)=20-4.9(1)^2=20-4.9\cdot1=20-4.9=15.1[/tex]When r = 1.1 seconds, we have:
[tex]H(1.1)=20-4.9(1.1)^2=20-4.9\cdot1.21=20-5.929=14.071\approx14[/tex]When r = 1.2 seconds, we have:
[tex]H(1.2)=20-4.9(1.2)^2=20-4.9\cdot1.44=20-7.056=12.944\approx13[/tex]To know the time for each height, we have to work with the equation like this:
[tex]\begin{gathered} H=20-4.9r^2 \\ 4.9r^2=20-H \\ r^2=\frac{20-H}{4.9} \\ r=\sqrt{\frac{20-H}{4.9}} \end{gathered}[/tex]Then, we can calculate at which time the height is 15 meters:
[tex]r=\sqrt{\frac{20-15}{4.9}=\sqrt{\frac{5}{4.9}}}=\sqrt{1.02}\approx1.01[/tex]When the height is 10 meters, r is:
[tex]r=\sqrt{\frac{20-10}{4.9}=\sqrt{\frac{10}{4.9}=}}\sqrt{2.04}\approx1.43[/tex]When the height is 5 meters, r is:
[tex]r=\sqrt{\frac{20-5}{4.9}=\sqrt{\frac{15}{4.9}=}}\sqrt{3.06}\approx1.75[/tex]The rock hits the ground when H=0. This happens when r is:
[tex]r=\sqrt{\frac{20-0}{4.9}=\sqrt{\frac{20}{4.9}=}}\sqrt{4.08}\approx2.02[/tex]