A projectile is fired straight upward from Earth's surface with a speed that is half the escape speed.If R is the radius of Earth, the highest altitude reached, measured from the surface, is:
A)R/4
B)R/3
C)R/2
D)R
E)2R

Question

Quizplus · Accepted Answer

The escape speed $v_e$ from Earth is given by the formula $v_e = \sqrt{2gR}$, where $g$ is the acceleration due to gravity and $R$ is the radius of Earth. When a projectile is fired with a speed $v$ that is half the escape speed, $v = \frac{1}{2}v_e = \frac{1}{2}\sqrt{2gR}$, the energy conservation principle can be used to find the maximum altitude $h$ it reaches. At the maximum altitude, the kinetic energy of the projectile is zero, and all the initial kinetic energy has been converted into potential energy.The initial kinetic energy $K_i$ of the projectile is $\frac{1}{2}mv^2$, and the potential energy at the surface of Earth is $-\frac{GmM}{R}$, where $m$ is the mass of the projectile, $M$ is the mass of Earth, and $G$ is the gravitational constant. At the highest point, the potential energy $U_f$ is $-\frac{GmM}{R+h}$, and the kinetic energy is zero.Setting the initial total energy equal to the final total energy gives:$$\frac{1}{2}mv^2 - \frac{GmM}{R} = 0 - \frac{GmM}{R+h}$$Substituting $v = \frac{1}{2}\sqrt{2gR}$ and simplifying, we find that $h = R/3$. This result comes from solving the equation for $h$ in terms of $R$, taking into account that $g = \frac{GM}{R^2}$.

A Projectile Is Fired Straight Upward from Earth's Surface with a Speed