Question

A ball is dropped from a top of a building at t=0. At a later time t=t0, a second ball is thrown downward with an initial speed up. What is the time at which the two balls meet?

Answer 1

time at which two balls meet is $\frac{gt_0^2}{2(u-gt_0)}$

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)

= $\frac{gt_0^2}{2(u-gt_0)}$

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Answer 2

$\huge\bold\purple{Answer:-}$

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)

= \frac{gt_0^2}{2(u-gt_0)}