Applied Calc- If an object is launched straight up into the air from a starting height of h_{0} feet, then the height of the object after t seconds is approximately h=-16t^2+v_{0}t+h_{0} feet, where v_{0} is the inital velocity of the object. Find the starting height and inital velocity of an object that attains a maximum height of 412 feet five seconds after being launched.

Question

Applied Calc- If an object is launched straight up into the air from a starting height of h_{0} feet, then the height of the object after t seconds is approximately h=-16t^2+v_{0}t+h_{0} feet, where v_{0} is the inital velocity of the object.  Find the starting height and inital velocity of an object that attains a maximum height of 412 feet five seconds after being launched.

anonymous · Answer

$$h(t)=-16t^2+v_{0}t+h_{0}$$
since max occurs at vertex which is $-\frac{b}{2a}=\frac{v_0}{32}=5$ we can solve for $v_0$

anonymous · Answer

we get $v_0=160$ and now you know 
$$h(5)=412=-16\times 25+160\times 5+h_0$$
$$412=-400+800+h_0$$
and we can solve for $h_0$

anonymous · Answer

i guess $h_0=12$

anonymous · Answer

didn't really use any calc, did we?

anonymous · Answer

i completley forgot about the vertex formula, thank you so much