Question

A ball is dropped from the top of a building at equals to zero at a later time T = 2 T not a second ball is thrown downward with initial speed V not obtain an expression for the time T at which the two balls meet

Answer 1

Answer:

time at which two balls meet is  

let height of building is h.

A ball is dropped from a top of a building at t = 0.

now another ball after time t = t0, thrown vertically downward with speed u.

so, distance covered by first ball during time t0 is , s = 1/2 gt0² ........(1)

velocity of first ball at t = t0, v = 0 + gt0 = gt0 ........(2)

now time at which the two balls meet, s = seperation between them/relative velocity

= s/(u - v)

= (1/2gt0²)/(u - gt0)

=