physics problem: integrals of a(t) v(t) s(t)

[pinkmongoose]

two balls are thrown upward from the edge of a cliff 432 ft above the ground. the first is thrown with a speed of 48ft/s and the other is thrown a second later with a speed of 24 ft/s. do the balls ever pass each other?

 

[soroban]

Hello, pinkmongoose!

No integrals are needed . . .

Two balls are thrown upward from the edge of a cliff 432 ft above the ground.
The first is thrown with a speed of 48 ft/s
and the other is thrown a second later with a speed of 24 ft/s.
Do the balls ever pass each other?

\(\displaystyle \text{Formula: }\;y \;=\;h_o + v_ot - 16t^2\quad \text{where: }\;\begin{Bmatrix} h_o &=& \text{initial height} \\ v_o &=& \text{initial velocity} \end{Bmatrix}\)

\(\displaystyle \text{For the first ball, the height is: }\;y_1 \;=\;432+48t - 16t^2\)

\(\displaystyle \text{The second ball, whose flight time is one second less, has height: }\;y_2 \;=\;432+24(t-1) - 16(t-1)^2\)


\(\displaystyle \text{If they pass each other, their heights are equal: }\;\;432+48t - 16t^2 \;=\;432+24(t-1) - 16(t-1)^2\)

. . \(\displaystyle \text{which simplifies to: }\;8t-40\:=\:0 \quad\Rightarrow\quad t \:=\:5\)


\(\displaystyle \text{Therefore, the balls will pass each other 5 second after the first ball is thrown.}\)
. . \(\displaystyle \text{They will be 272 feet above the ground.}\)

 

[pinkmongoose]

Thanks! I was supposed to solve the problems using integration i.e deriving the formulas for motion as you pointed them out, but thanks to that I understand what I'm supposed to get