Sloth,

You're answer is incorrect because in your first step you are confusing the t^2 term in the position function with the t term in the velocity function. In your first step you incorrectly used t^2 rather than t, which causes your units for the answer to come out wrong. In your version after solving for t and paying close attention to units the answer comes out to t=9.035 (seconds^(1/2)) Or, in simpler terms the units are the square root of seconds. To get the desired units we'd have to square the answer only to get t=81.6 seconds.

The vertical position as a function of time is a parabola, implying a quadratic function which is given as h(t)=-4.9t^2-800t.

The reason for this is given in my explanation above and involves working backward to find the position function by knowing that the velocity function v(t) is the derivative of the position function h(t), and that the acceleration function a(t) is the second derivative of h(t), or the derivative of v(t). Working backward as I explained above gives us the following functions that describe the motion of the bullet (sans drag) as a function of time t.

Acceleration a(t)=-9.8m/s^2 (this is a constant function because accelleration is contant regardless of time)

Velocity v(t)=A(t) (the antiderivative of a(t))=h'(t)= -9.8 (m/s^2) X t (sec.) + 800(m/s) if you work this out you'll see that the units cancel out to give the answer in meters/sec., which is a unit of velocity that we're looking to find with this function.

Height (in meters m) h(t)=V(t)=-4.9(m/s^2) x t(s)^2 + 800(m/s) x t(s) + 0(m) (the last term is the starting position, which we're calling zero since we consider the muzzle the zero)

Solving the velocity function for zero gives us t=81.6 seconds when v(t)=0.

Plugging t=81.6seconds into the height function h(t) we find that the height in meters when velocity is zero is 32,653 meters [h(81.6)=32,653m].

Mike