Question

A projectile is launched vertically from the surface of the Moon with an initial speed of 1360 m/s. At what altitude is the projectile's speed two-fifths its initial value?

Answer 1

Answer:

485520 m

Explanation:

$v_{o}$ = initial velocity of the projectile = 1360 m/s

$v_{f}$ = final velocity of the projectile = $\left ( \frac{2}{5} \right )v_{_{o}}$ = $\left ( \frac{2}{5} \right )(1360)$ = 544 m/s

a = acceleraton due to gravity on moon = - 1.6 m/s²

h = Altitude of the projectile

Using the kinematics equation

$v_{f}^{2} = v_{o}^{2} + 2 a h$

Inserting the values

$544^{2} = 1360^{2} + 2 (-1.6) h$

h = 485520 m