A ball is dropped from a height h and falls the last half of its distance in 4 seconds. How long does the ball fall? From what height is the ball originally dropped? Any help is greatly...

A ball is dropped from a height and falls the last half of its distance in 4 seconds.

How long does the ball fall?

From what height is the ball originally dropped?

Any help is greatly appreciated!

Asked on by mbell2017

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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

Posted on

Hello!

The height of the ball as a function of time t is

`H(t) = h - (g*t^2)/2` ,

where h is the initial height and g is the gravity acceleration.
I suppose that the initial velocity is zero and the air resistant is ignored.
And that the ball ends its fall at the height 0.

Denote the time when the ball is at the half of the initial height as `t_1` and the time when the ball hits the floor as `t_2` (in seconds).

`H(t_1) = h/2,`

`H(t_2) = 0.`

It is given that `t_2-t_1 = 4` . Let's find `t_1` and `t_2` in terms of h:

`h - (g*t_1^2)/2 = h/2,`
`h - (g*t_2^2)/2 = 0.`

This implies
`(g*t_1^2)/2 = h/2,`
`(g*t_2^2)/2 = h`

and further
`t_1 = sqrt(h/g),`
`t_2 = sqrt(2h/g) = sqrt(2)*sqrt(h/g).`

Now we can find h:
`t_2 - t_1 =sqrt(2)*sqrt(h/g) -sqrt(h/g) = (sqrt(2)-1)*sqrt(h/g) = 4.`

From this
`sqrt(h/g) = 4/(sqrt(2)-1) = 4*(sqrt(2)+1)`

and finally
`h = g*16*(sqrt(2)+1)^2.`

g is about 9.8m/s^2, therefore
`h = 9.8*16*(sqrt(2)+1)^2 approx 913.90 (m).`

Now we can find t_2,
`t_2 = sqrt(2h/g) = sqrt(2*913.9/9.8) = sqrt(186.51) approx 13.66 (s).` 

How long does the ball fall is `t_2 =`  13.66 (s).

From what height is the ball originally dropped is `h=`  913.90 (m).

For the visualisation, please look at the graph:
https://www.desmos.com/calculator/k9l2jo3x6k

(Actually, for these heights the air resistance cannot be ignored.
But to take the air resistance into account we also have to know the mass of the ball and its size, not speaking about its shape.)

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