Answer:
(a) t = 2.97s
(b) h = 43.3 m
Explanation:
Let t be the time it takes to fall a distance h, then t - 1 (s) is the time it takes to fall a distance of h - 0.56h = 0.44 h
For the ball to fall from rest a distance of h after time t
[tex]h = gt^2/2[/tex]
Also for the ball to fall from rest a distance of 0.44h after time (t-1)
[tex]0.44h = g(t-1)^2/2[/tex]
We can substitute the 1st equation into the 2nd one
[tex]0.44gt^2/2 = g(t-1)^2/2[/tex]
and divide both sides by g/2
[tex]0.44t^2 = (t-1)^2[/tex]
[tex]0.44t^2 = t^2 - 2t + 1[/tex]
[tex]0.56t^2 - 2t + 1 = 0[/tex]
[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]t= \frac{2\pm \sqrt{(-2)^2 - 4*(0.56)*(1)}}{2*(0.56)}[/tex]
[tex]t= \frac{2\pm1.33}{1.12}[/tex]
t = 2.97 or t = 0.6
Since t can only be > 1 s we will pick t = 2.97 s
(b) [tex]h = gt^2/2 = 43.3 m[/tex]
