Question

An object falls a distance h from rest. If it travels 0.560h in the last 1.00 s, find (a) the time and (b) the height of its fall.

Answer 1

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

$h = gt^2/2$

Also for the ball to fall from rest a distance of 0.44h after time (t-1)

$0.44h = g(t-1)^2/2$

We can substitute the 1st equation into the 2nd one

$0.44gt^2/2 = g(t-1)^2/2$

and divide both sides by g/2

$0.44t^2 = (t-1)^2$

$0.44t^2 = t^2 - 2t + 1$

$0.56t^2 - 2t + 1 = 0$

$t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$t= \frac{2\pm \sqrt{(-2)^2 - 4*(0.56)*(1)}}{2*(0.56)}$

$t= \frac{2\pm1.33}{1.12}$

t = 2.97 or t = 0.6

Since t can only be > 1 s we will pick t = 2.97 s

(b) $h = gt^2/2 = 43.3 m$