Two students are on a balcony a distance h above the street. One student throws a ball vertically downward at a speed vi; at the same time, the other student throws a ball vertically upward at the same speed. Answer the following symbolically in terms of vi, g, h, and t. (Take upward to be the positive direction.) (a) What is the time interval between when the first ball strikes the ground and the second ball strikes the ground?
?t =
(b) Find the velocity of each ball as it strikes the ground.
For the ball thrown upward vf =
For the ball thrown downward vf =
(c) How far apart are the balls at a time t after they are thrown and before they strike the ground?
d =

Explanation:

 a )

Time for first ball to reach top position

v = u - gt

0 = vi - gt

t = vi / g

Time to reach balcony  while going downwards

= vi /g

Total time = 2 vi / g

Time to go down further to the ground = t₁

Total time = 2 vi / g + t₁

Time for the other ball to go to the ground = t₁

Time difference = ( 2 vi / g + t₁ ) - t₁

= 2vi / g .

( b )

v² = u² + 2gh

For both the throw ,

final displacement = h , initial velocity downwards = vi

( For the first ball also  , when it go down while passing the balcony , it acquires the same velocity vi but its direction is downwards.)

vf² = vi² + 2gh

vf = √ ( vi² + 2gh )

(c )

displacement of first ball after time t

s₁ = - vi t + 1/2 g t²  [ As initial velocity is upwards , vi is negative ]

displacement of second ball after time t

s₂ = vi t + 1/2 g t²

Difference = d =  s₂ - s₁

= vi t + 1/2 g t² - ( - vi t + 1/2 g t² )

d = 2 vi t .

a. are

this could be newton#'s law of cooling. the cooling curve is exponential