Question 207888
{{{h = - 16 t^2 + 40 t + 24}}}
{{{h = -8*(2t^2 - 5t - 3)}}}
{{{h = -8*(2t+1)(t-3)}}}

You can see that at t=3, the term (t-3) above would evaluate to 0. 
Since 0* anything = 0, then h(3) = 0.