Finding minimum distance between two particles ( kinematics)

Question

A particle is dropped from the building of height 100m and another particle was projected at an angle 60° with the horizontal with intial velocity u=60m/s at the same time. Find the minimum distance between two particles. 

Answer 1

The position of the falling particle is:

 

x = 0

y = 100-9.8t²

 

The position of the projected particle is:

 

x = (60cos(60))t = 30t

y = (50sin(60))t -9.8t²= 30sqrt(3)t-9.8t²

 

Think of the distance between the 2 particles as an ever changing right triangle where y = height of the first particle minus the height of the second particle and x is the length the 2nd particle has traveled away from the building.  So distance equals:

 

sqrt[(30t)² + (100-30sqrt(3)t)²]

 

simplify and get

 

(900t² + 10000 - 6000sqrt(3)t + 2700t²)^1/2 = (3600t² - 6000sqrt(3)t + 10000)^1/2

 

Take the first derivative, set it equal to zero and solve for t:

 

1/2(3600t² - 6000sqrt(3)t + 100000)^-1/2(7200t - 6000sqrt3) = 0

 

Since the denominator cannot be zero, we get:

 

7200t - 6000sqrt(3) = 0

 

t = 6000sqr(3)/7200 = 1.44 seconds

Answer 2

Where does the second particle originate? At the bottom of the building?