# 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 so that it takes 3 s to strike the floor? 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...

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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.

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