Question

A rock is dropped off a cliff and falls the first half of the distance to the ground in t1 seconds. If it falls the second half of the distance in t2 seconds, what is the value of t2/t1? (Ignore air resistance.)

Answer 1

Answer:

$t2/t1 = \sqrt{2} - 1$

Explanation:

The expression for the second law of motion is given below:

h = $ut + 0.5at^2$

For first half distance

Object is initially at rest, so its initial speed u = 0

Object falls at half the distance, so h = h/2 where t = t1

Hence, we have

$h/2 = at1^2/2 - equation 1$

For second half distance:

Similarly,

$h = a(t1 + t2)^2/2 - equation 2$ where t = t1 + t2 and u= 0

Using equation 2 by equation 1

we obtain $2 = (t1 + t2)^2/t1^2$

Hence $t2/t1 + 1 = \sqrt{2}$

Hence $t2/t1 = \sqrt{2} - 1$