Question

(3rd time Asking) Please help me with these questions!!!

Answer 1

(a) Water balloon 1 has height at time $t$

$h_1(t)=-16t^2-20t+110$

and water balloon 2 has height

$h_2(t)=-16t^2+75$

Both water balloons are moving straight downward, so their distance is equal to

$|h_1(t)-h_2(t)|=|(-16t^2-20t+110)-(-16t^2+65)|=|35-20t|$

(b) When both water balloons are at the same height, we have $h_1(t)=h_2(t)$. So you could set both quadratic expressions equal to each other and solve for $t$. 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

$|35-20t|=0\implies 35-20t=0\implies 35=20t\implies t=\dfrac{35}{20}$

or 1.75 seconds after being thrown/dropped.

At this time, they are both at a height of $h_1(1.75)=h_2(1.75)$:

$h_2(1.75)=-16(2.25)^2+75=26$

feet.

(c) The second and fourth statements are true.