Question

A projectile is launched vertically from the surface of the Moon with an initial speed of 1460 m/s. At what altitude is the projectile's speed one-half its initial value?

Answer 1

Given,

The initial speed of the projectile, $u\ =\ 1460\ m/s$

The final speed of the projectile, $v\ =\ \dfrac{u}{2}$

  $=\ \dfrac{1460}{2}\ m/s$

  $=\ 730\ m/s$

Acceleration due to gravity on the surface of the moon, $g\ =\ 1.625\ m/s$

Since the projectile is going up so there will be de-acceleration in the projectile, hence from the equation of motion,

$v^2\ =\ u^2-2gh$

where h is the height reached by the projectile.4

$h\ =\ \dfrac{u^2-v^2}{2g}$

  $=\ \dfrac{(1460)^2-(730)^2}{2\times 1.625}$

  $=\ 491907.69\ m$

  $=\ 491.907\ km$

Hence, the altitude at which the projectile's speed is one-half of its initial value is 491.907 km.