a cannonball is launched horizontally from the top of an 78.4-meter high cliff. how much time will it take for the ball to reach the ground

Do you want me to use calculus or physics formulas?

Calculus:

Initially  

 $$\\\ddot y=-9.8\;m/s^2, \qquad \dot y=0\;m/s, \qquad y=78.4\;m \\\\
\mbox{Find t when y=0}\\\\$$

Ongoing:

$$\\\ddot y=-9.8\;m/s^2\\\\
\ddot y=-9.8t+C_1 \\\\
c_1 =0 \qquad \mbox{because of initial conditions therefore} \\\\
\dot y=-9.8t\\\\
y=\frac{-9.8t^2}{2}+C_2\\\\
C_2 =78.4 \qquad \mbox{because of initial conditions therefore} \\\\
y=-4.9t^2+78.4$$

find t when y=0

$$\\0=-4.9t^2+78.4\\\\
-78.4=-4.9t^2\\\\
16=t^2\\\\
t=4\;seconds$$

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If you want to do it by physics formula you will use         $$s=ut+\frac{1}{2}at^2$$

and you will get the same result.