when h=0, that is when it hits the ground
0=70-4t-16t²
based on your previous questions, you should have learned how to complete the square
solve by copmleting the square
minus 70 both sides
-70=-4t-16t²
times both sides by -1
70=4t+16t²
16t²+4t=70
divide both sides by 16
t²+1/4t=35/8
take 1/2 of the linear coefint and square it
half of 1/4=1/8, (1/8)²=1/64
add that to both sides
t²+1/4t+1/64=35/8+1/64
factor perfect square trionomial
[tex](t+\frac{1}{8})^2=\frac{35}{8}+\frac{1}{64}[/tex]
[tex](t+\frac{1}{8})^2=\frac{280}{64}+\frac{1}{64}[/tex]
[tex](t+\frac{1}{8})^2=\frac{281}{64}[/tex]
square root both sides, remember to take positive and negative roots
[tex]t+\frac{1}{8}=\frac{+/-\sqrt{281}}{8}[/tex]
minus 1/8
[tex]t=\frac{-1+/-\sqrt{281}}{8}[/tex]
[tex]t=\frac{-1+\sqrt{281}}{8}[/tex] or [tex]t=\frac{-1-\sqrt{281}}{8}[/tex]
the 2nd solution is not valid because that gives us a negative time, before the ball was dropped
so [tex]t=\frac{-1+\sqrt{281}}{8}[/tex] is the solution
t≈1.97
1.97 seconds
