Question

Integrals Question With approximately what velocity does a diver enter the water after diving from a 30-m platform? (Use g=9.8m/sec). Thank you!!!

Answer 1

`~~ 24.2 \ ms^-1`

Explanation:

An object free falling under the act of gravity alone falls with constant acceleration, `g`. As acceleration is the rate of change of velocity, `v`, wrt time we can write:

`(dv)/dt = g` ..... [A]

This is a First Order, Separable ODE, and we can directly integrate [A] to get

`v = g t + c_1`

Assuming that the diver "falls" from the platform (rather than jumps), so that their initial velocity is zero, then we have an initial condition:

`v=0` when `t=0 => 0=0+c_1` so that `c_1 = 0`

So, we can write:

`v = g t` ..... [B]

Equivalently, as velocity is the rate of change of displacement `x`, wrt time, then we have:

`dx/dt = g t` ..... [C]

Again, this is a First Order, Separable ODE, and we can directly integrate [C] to get

`x = 1/2g t^2 + c_2`

If we consider the relative displacement, then initially there is n o movement of the diver leading to a further initial; condition:

`x=0` when `t=0 => 0 = 0 + c_2` so that `c_2 =0`

So, we can write:

`x = 1/2g t^2`

If if takes a time `T`, for the diver to fall a relative displacement of `30 \ m`, then we have

`30 = 1/2g T^2 => T = sqrt(60/g) \ s`

And then using the earlier equation [B], we can determine the instantaneous velocity at this time:

`v_30 = g T = gsqrt(60/g)`

If we use `g=9.8`, as indicated, then the velocity we seek is thus:

`v_30 = 9.8sqrt(60/9.8)`
`\ \ \ \ \ = 9.8 * 2.47435 ...`
`\ \ \ \ \ = 24.248711...`
`\ \ \ \ \ ~~ 24.2 \ ms^-1`