What is the output of a signal in time domain that passed through a High Pass Filter with simple transfer function

What you need to do is

1. use Laplace transform to $U(t)$ so you would get $mathcal{L}(U(t))=U(s)$

for example, if you are rusty on Laplace transforms (or their inverse), you can use wolfram alpha

2. multiply $H(s)*U(s)$

that will get you a function of s $Y(s)$

3. use the inverse laplace transform to Y(s) to get the $y(t) = mathcal{L}^{-1}(Y(s))$

Luckily in this case there is a symbolic result.

OutputResponse[TransferFunctionModel[s/(s + w), s], 25 - (5 - t)^2, t]
res1 = Simplify[%[[1]]]

$-frac{2 e^{-t w} left(e^{t w} ((t-5) w-1)+5 w+1right)}{w^2}$

A procedure to get this (as mentioned) is

s/(s + w) LaplaceTransform[25 - (5 - t)^2, t, s]
res2 = InverseLaplaceTransform[%, s, t]

$-frac{2 e^{-t w} left(e^{t w} ((t-5) w-1)+5 w+1right)}{w^2}$

The results for various values of $w$

Table[w, {w, 0.1, 1, 0.2}];
Plot[Evaluate@Table[res, {w, %}], {t, 0, 10}, PlotLegends -> %]