Launching a ball on a tiny planet

Let $M$ be the mass of the planet and $m$ be the mass of the projectile. Newton's law of universal gravitation states that

$F = G\frac {Mm}{r^2}$

The work required to move the projectile over a small distance $dr$ is

$dW = F dr = -G\frac {Mm}{r^2} dr$

The minus sign is to indicate that the work is being done against gravity.

The total work needed to raise the projectile to a distance $r_h$ away from the planet's center is

$W=\int_{r_0}^{r_h} -G\frac {Mm}{r^2} dr = GMm(\frac{1}{r_h} - \frac{1}{r_0})$

where $r_0$ is the radius of the planet.

Now, $F = mg$ , so that $g = \frac{F}{m} = \frac{GM}{{r_0}^2}$

Therefore, $GM = g{r_0}^2$

By the work-energy principle,

$W + KE = 0 \implies GMm(\frac{1}{r_h} - \frac{1}{r_0}) + \frac{1}{2} m{v_0}^2 = 0$

$-g{r_0}^2(\frac{1}{r_h} - \frac{1}{r_0}) = \frac{1}{2}{v_0}^2$

Substituting the values from the problem gives us

$-(10m/s^2)(4m)^2(\frac{1}{r_h}-\frac{1}{4m}) = \frac{1}{2}(7.5m/s)^2$

$\frac{1}{r_h}-\frac{1}{4}m^{-1} = -\frac{45}{256}m^{-1}$

$\frac{1}{r_h} = \frac{19}{256}m^{-1}$

$r_h = 13.473684m$

But remember, $r_h$ is the maximum distance of the projectile from the center of the planet. To get the final answer, we subtract the planet's radius to get $\boxed{9.473684m}$ .