A ball initially at rest is dropped from a height y above the floor and it hits the floor after 1.5 s. From what height should the ball be dropped...

The equation relating height to time for an object in free fall is

`h = 1/2 g t^2`

where h is height, g is acceleration due to gravity and t is time. Since time is a squared factor of height, for the object to take twice as long to strike the floor it must be dropped from four times the height, or h = 4y.

It's easy enough to calculate the two heights to see this. I'll use 10 m/s^2 for acceleration due to gravity:

y = (1/2)(10 m/s^2)(1.5 s)^2 = 11.25 meters

second height: h = (1/2)(10 m/s^s)(3 sec) = 45 m

45/11.25 = 4, so the second height is four times y

You could also solve this by canceling out the factors that are the same in both scenarios, 1/2 g:

`(h_1)/(t_1)^2 = (h_2)/(t_2)^2`  so `(h_2)/(h_1) = (t_2)^2/(t_1)^2`

Therefore the factor by which the height changes is the square of the factor by which the time changes.