(a) Water balloon 1 has height at time [tex]t[/tex]
[tex]h_1(t)=-16t^2-20t+110[/tex]
and water balloon 2 has height
[tex]h_2(t)=-16t^2+75[/tex]
Both water balloons are moving straight downward, so their distance is equal to
[tex]|h_1(t)-h_2(t)|=|(-16t^2-20t+110)-(-16t^2+65)|=|35-20t|[/tex]
(b) When both water balloons are at the same height, we have [tex]h_1(t)=h_2(t)[/tex]. So you could set both quadratic expressions equal to each other and solve for [tex]t[/tex]. Or you can use the fact that, when they're at the same height, the distance between them is 0. So using the expression from part (a), we have
[tex]|35-20t|=0\implies 35-20t=0\implies 35=20t\implies t=\dfrac{35}{20}[/tex]
or 1.75 seconds after being thrown/dropped.
At this time, they are both at a height of [tex]h_1(1.75)=h_2(1.75)[/tex]:
[tex]h_2(1.75)=-16(2.25)^2+75=26[/tex]
feet.
(c) The second and fourth statements are true.
