Question

A projectile is launched vertically from the surface of the Moon
with an initial speed of 1100 m/s. At what altitude is the
projectile's speed one-fourth its initial value?

Answer #1

**Given**

** projectile launched vertiallcy from surface
of Moon where**

**acceleration due to gravity g = 1.633 m/s2**

** with initial velocity u = 1100
m/s**

**and the final velocity v = u/4**

** height h = ?**

**from equations of motion**

** v^2 - u^2 = -
2*g*h**

** h = v^2 - u^2 /
(-2*g)**

**substituting
h = (1/4 u)^2 -u^2 /(-2*g)
h = u^2(1/16 -1) /2g
h = 1100^2(-15/16) /(-2*1.633)
m
h = 347328.53644 m**

**so the altitude where the velocity becomes 1/4 th of
initial velocity is 347328.53644 m**

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