Question

A ball is tossed with enough speed straight up so that it is in the air serveral seconds. Assume upwards direction is positive and downward is negative. What is the change in its velocity during this 1-s interval?

Answer 1

- 32.17 fts/s in the imperial system, -9.8 m/s in the SI

Step-by-step explanation:

We know that acceleration its the derivative of velocity with respect to time, this is (in 1D):

`a = (dv)/(dt)`

So, if we wanna know the change in velocity, we can take the integral:

`v(t_f) - v(t_i) = \int\limits^(t_f)_(t_i) {(dv)/(dt) \, dt = \int\limits^(t_f)_(t_i) a \, dt`

Luckily for us, the acceleration in this problem is constant

`a \ = \ - g`

the minus sign its necessary, as downward direction is negative. Now, for a interval of 1 second, we got:

`t_f = t_i + 1 s`

`v(t_i + 1 s) - v(t_i) =  \int\limits^(t_i + 1 s)_(t_i) a \, dt`

`v(t_i + 1 s) - v(t_i) =  \int\limits^(t_i + 1 s)_(t_i) (-g ) \, dt`

`v(t_i + 1 s) - v(t_i) =  [-g t]^(t_i + 1 s)_(t_i)`

`v(t_i + 1 s) - v(t_i) =  -g (t_i+1s) + g t_i`

`v(t_i + 1 s) - v(t_i) =  -g t_i + -g 1s + g t_i`

`v(t_i + 1 s) - v(t_i) =  -g 1s`

taking g in the SI

`g=9.8 (m)/(s^2)`

this is:

`v(t_i + 1 s) - v(t_i) =  - 9.8 (m)/(s)`

or, in imperial units:

`g=32.17 (fts)/(s^2)`

this is:

`v(t_i + 1 s) - v(t_i) =  - 32.17 (fts)/(s)`