The maximum height reached by a projectile projected vertically upwards with speed equal to half the escape velocity ($v

Question

The maximum height reached by a projectile projected vertically upwards with speed equal to half the escape velocity ($v = \frac{ve}{2}$) from Earth's surface is:

Accepted Answer

By conservation of energy: $-\frac{GMm}{R} + \frac{1}{2}m v^2 = -\frac{GMm}{R+h}$. Since $v^2 = \frac{ve^2}{4} = \frac{2GM}{4R} = \frac{GM}{2R}$, substituting gives $h = \frac{R}{3}$.

Gravitation

❓ Sibling Questions in this Chapter