A body falls vertially from a height

Question

A body with an initial velocity of zero falls vertically from a height of h = 19.6 m. How long it takes for the body to go: 1) its paths first 1 m; 2) its paths last 1 m? Ignoring air resistance. [1] 0.45 s; 2) 2 s; 0.05 s].

Answer 1

The change of position is the product of the initial velocity and time summed with half the product of the acceleration and the square of the time:

∆y = Vi•t + 1/2 a•t²

The initial velocity is zero: Vi = 0

∆y = 0•t + 1/2 a•t² = 0 + 1/2 a•t² = 1/2 a•t²

The acceleration of gravity near the surface of the earth is about 9.8 m/s². The height for the 1st meter is ∆y = h = 1 m ∴ re-arrange the equation to define time in terms of acceleration and height.

∆y = 0•t + 1/2 a•t² = 0 + 1/2 a•t² = 1/2 a•t² = h

1/2 a•t² = h

a•t² = 2h

t² = 2h/a

t = (2h/a)^1/2

Plug in known values

t = (2•1/9.8)^1/2 = 4^1/2 = 0.45 sec

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The time it takes to fall the last meter is the difference between the time it takes to fall 19.6m and a 18.6 m...

The acceleration of gravity near the surface of the earth is about 9.8 m/s². ∆y is the height - ∆y = h = 19.6m ∴ re-arrange the equation to define time in terms of acceleration and height.

We can use the same equation:

...

t = (2h/a)^1/2

Plug in known values

t = (2•19.6/9.8)^1/2 = 4^1/2 = 2sec

To fall 18.6 meters is:

t = (2•18.6/9.8)^1/2= 1.95 sec

So it takes just 2s-1.95s= .05 sec to travel the last meter!