How long does it take a ball to reach the ground when dropped from the top of a 484 foot building?  

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Hello!

We'll ignore the air resistance. Also I suppose that "dropped" means that the initial speed is zero.

 

Then the speed `V` of a ball increases uniformly in magnitude under the constant gravity force, `V=-g*t` (with respect to an upward axis).

The height `H(t)` is `H_0-g*(t^2)/2`  where `H_0` is the initial height and `g=32 ft/s^2` is the gravity acceleration (I think you know this). `t` is for time in seconds.

 

A ball reaches the ground when its height is zero, so we have to solve the equation `H(t)=0` for `tgt0.` It is simple:

`H_0-g*(t^2)/2=0,`  `H_0=g*(t^2)/2,`  so `t=sqrt(2H_0/g) =5.5 (s).` This is the answer.

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