EDIT: It gets the same answer.
I have read that the time scaling property of the Laplace transform is not relevant for the unit step function. This property being: $$\mathcal L \left[f(at)\right]= \left(\frac{1}{a}\right)F\left(\frac{s}{a}\right)$$ Does this mean that: $$\mathcal L \left[5u\left(\frac{t}{2}\right)\right]= \left(\frac{5}{s}\right)$$ Or does it still hold and the transform is then: $$\mathcal L \left[5u\left(\frac{t}{2}\right)\right]= 5\left(\frac{1}{\left(\frac{1}{2}\right)}\right)\left(\frac{1}{\left(\frac{s}{\left(\frac{1}{2}\right)}\right)}\right)=\left(\frac{10}{1}\right)\left(\frac{1}{2s}\right) = \left(\frac{5}{s}\right)$$