Question 879720
1) The velocity is {{{highlight(64-32t)}}} .
2) You do not need to solve for {{{H}}} ,
because you have the formula {{{H=64t-16t^2}}} to calculate {{{H}}} when you are given {{{t}}} .
3) To solve for {{{t}}} you have to solve a quadratic equation.
{{{H=64t-16t^2}}}<-->{{{16t^2-64t+H=0}}} ,
and the solutions to {{{16t^2-64t+H=0}}} , if there is any,
can be calculated applying the quadratic formula as
{{{t = (64 +- sqrt(64^2-4*16*H ))/(2*16) }}}-->{{{t = (64 +- sqrt(64^2-64H) )/32}}}-->{{{t = (64 +- sqrt(64(64-H)))/32}}}-->{{{t = (64 +- 8sqrt(64-H))/32}}}-->{{{highlight(t = (8 +- sqrt(64-H))/4)}}}
That could give two answers, or one, or none.
For example, {{{H=0}}} when {{{t=0}}} and when {{{t=4}}} ;
{{{H=64}}} for {{{t=2}}} only,
and {{{H=70}}} does not happen for any real value of {{{t}}} .
 
The explanations for the answers above depend on why you need the answers, and what courses you have taken and/or are taking. It is different for Calculus than for Algebra 2 or Pre-calculus. It is different for Physics with little math.
 
For Physics, you should know that {{{H=64t-16t^2}}} represents
the height in feet after {{{t}}} seconds
of an object shot up from the ground with an initial upwards velocity of {{{64}}} feet per second,
on Earth, which pulls down on objects with an acceleration of {{{32}}} {{{ft}}}{{{"/"}}}{{{s^2}}} .
 
For Calculus, I would tell you that the velocity is the derivative of {{{H(t)}}} , {{{dH/dt}}} .
 
I would give more explanations if I knew which ones you need and can use.