Finding maximum height with initial velocity. 
An object is projected from the surface of Earth 
  with a velocity of one mile per second. What is 
  the maximum height it will reach?

My solution for solving this was to use the following equation:
$$v = \sqrt{V^2+2GM\biggl(\frac{1}{R} - \frac{1}{r_1}\biggl)}$$
This equation (derived using calculus) gives the velocity of an object with an initial velocity of $V$ and where $GM$ is the quantity $32(4000^2 * 5280^2)$ and $R$ is the radius of the Earth in feet (approximated as $4000*5280$ feet). $r_1$ is the height of the object from the centre of the Earth and is the value I'm solving for. The equation basically accounts for the force of gravity on an object as it gets further or closer to Earth.
When I plugged $5280$ ft/s for $V$ into this formula along with the values of the other variables, I obtained approximately 3919 miles from the center of the Earth. This value doesn't make sense since if we use the value 4000 miles as the radius of the Earth, then the result will be a negative value when subtracted. 
I then assumed that the gravity of the Earth is constant, and I solved the equation $$v = 5280 - 32t$$ for $t$ when $v$ = 0. I then plugged that into the derived formula for distance $d = 5280t - 16t^2$ and I obtained 82.5 miles as the maximum height. The book I'm using says the answer is 84 miles.
My mind is blanking and I'm unable to determine what I exactly can do to get close to 84 miles. If anyone can help me, that would be greatly appreciated.
Thanks.
 A: I'm not sure exactly how your equation was explained to you, but it is based on conservation of energy. The sum of kinetic and gravitational potential energy of a particle in the Earth's gravitational field is
$$ E = \frac12 mv^2 - \frac{GMm}{r}.$$
Assuming your particle is in a ballistic trajectory, ignoring atmospheric resistance, this quantity is constant throughout the trajectory, in particular it is the same at the start and at its highest point. That is, denoting the speed and radius at the start as $v_0, r_0$ and at the end as $v_1, r_1$,
$$
\frac12 mv_0^2 - \frac{GMm}{r_0} = \frac12 mv_1^2 - \frac{GMm}{r_1}.
$$
Multiplying through by $2/m$ and "moving" a term to the right side we have
$$
v_0^2 =  v_1^2 + \frac{2GM}{r_0} - \frac{2GM}{r_1}
 =  v_1^2 + 2GM\left(\frac{1}{r_0} - \frac{1}{r_1}\right).
$$
We get something like your equation by taking the square root of both sides.
But let's not bother with the square root; I think it's a nuisance.
Let's just take $v_0 = 5280,$
$r_0 = 4000\cdot 5280,$ $v_1 = 0,$ and $GM = 32(4000\cdot 5280)^2,$
and plug them in, with the result
$$
5280^2 =  0 + 2\cdot32(4000\cdot 5280)^2\left(\frac{1}{4000\cdot 5280} - \frac{1}{r_1}\right).
$$
When you solve for $r_1$ in this equation you'll get an answer greater than $4000$ miles.
If $R$ is the radius of the Earth, the equation as you wrote it is correct when $v$ (on the left side of the equation) is the initial speed and $V$ is the speed upon reaching radius $r_1$ from the center of the Earth.
It looks like you misapplied the formula by substituting your initial and final speeds in the wrong places. 
A: Your answer is correct. I don't know what's wrong, but your book might use a different value of $g$.
Using another formula:
$$H = \frac{{v_0}^2 \sin^2 \theta}{2g}$$
where $H$ is the maximum height, $v_0$ is the initial velocity, $\theta$ is the launch angle, and $g$ is the acceleration due to gravity.
However, $g \approx 9.8 \ \text{m}/\text{s}^2$. Converting it into $\text{mi}/\text{s}^2$, we get $g \approx \frac{9.8}{1610} \ \text{mi}/\text{s}^2$.
The maximum of $\sin \theta$ in the range $[0, \pi]$ is $\theta = \frac{\pi}{2}$, so the maximum of $\sin^2 \theta$ in the same range is also $\theta = \frac{\pi}{2}$. Substituting this in, we have:
$$H = \frac{(1)^2 \sin^2(\frac{\pi}{2})}{2 \cdot \frac{9.8}{1610}}$$
$$= 82.1 \ \text{miles}$$
to $1 \ \text{d.p}$.
