# An object is dropped from a helicopter at a height of 59m above the ground. How long does it take for someone on the ground to catch it? Suppose the person catching the object is 2m tall.

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An object is dropped from a helicopter 59m from the ground. How long did it take to drop to a person on the ground?

We use the falling object model; the distance d, measured in meters, an object falls in time t, measured in seconds, is given by:

`d=-4.9t^2+v_0t+s_0 `

where v(0) is the initial velocity and s(0) is the initial height, and d the final height.

Here s(0)=59m and v(0)=0m/s. (We assume that the object was dropped, or released, with no downward push.)

When the object is caught, it will have fallen 57m (the person catches it 2m in the air) so d=57.

2=-4.9t^2+59

-57=-4.9t^2

t^2= 11.633

t is approximately 3.411 seconds.

So the object will be caught after t=3.4 seconds

Note -- some texts use -5m/s^s as the acceleration due to gravity instead of -4.9. Of course we are assuming that we are on Earth, we neglect air resistance, and all of the other typical simplifications.