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# According to Newton's Law of Universal Gravitation, the gravitational force on an object of mass $m$ that has been projected vertically upward from the earth's surface is$F = \frac {mgR^2}{(x + R)^2}$where $x = x(t)$ is the object's distance above the surface at time $t, R$ is the earth's radius, and $g$ is the acceleration due to gravity. Also, by Newton's Second Law, $F = ma = m (dv/dt)$ and so$m \frac {dv}{dt} = - \frac {mgR^2}{(x + R)^2}$(a) Suppose a rocket is fired vertically upward with an initial velocity $v_0.$ Let $h$ be the maximum height above

## a) $V_{o}=\sqrt{\frac{2 g R h}{h+R}}$b) $\sqrt{2 g R}$c) $36580 f t / s, 6.93 \mathrm{mi} / \mathrm{s}$

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